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Intermediate Exam II: Problem #1 (Spring 05) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the voltage V 3 across capacitor C 3 . (c) Find the the charge Q 2 on capacitor

  1. Unit Exam II: Problem #1 (Spring ’08) The circuit of capacitors is at equilibrium. (a) Find the charge Q 1 on capacitor 1 and the charge Q 2 on capacitor 2. (b) Find the voltage V 1 across capacitor 1 and the voltage V 2 across capacitor 2. (c) Find the charge Q 3 and the energy U 3 on capacitor 3. C 3 = 5 µ F Solution: « − 1 „ 1 1 C 1 = 6 µ F = 12 µ F (a) C 12 = 6 µ F + = 4 µ F , C 2 12 µ F Q 1 = Q 2 = Q 12 = (4 µ F)(12V) = 48 µ C . (b) V 1 = Q 1 = 48 µ C 12V 6 µ F = 8V , C 1 V 2 = Q 2 = 48 µ C 12 µ F = 4V . C 2 (c) Q 3 = (5 µ F)(12V) = 60 µ C , U 3 = 1 2 (5 µ F)(12V) 2 = 360 µ J . 1/5/2019 [tsl377 – 12/60]

  2. Unit Exam II: Problem #2 (Spring ’08) Consider the electric circuit shown. Find the current I 1 through resistor 1 and the voltage V 1 across it (a) when the switch S is open, (b) when the switch S is closed. (c) Find the equivalent resistance R eq of the circuit and the total power P dissipated in it when the switch S is closed. = 4Ω R 3 S = 4Ω R 1 = 2Ω R 2 12V 1/5/2019 [tsl378 – 13/60]

  3. Unit Exam II: Problem #2 (Spring ’08) Consider the electric circuit shown. Find the current I 1 through resistor 1 and the voltage V 1 across it (a) when the switch S is open, (b) when the switch S is closed. (c) Find the equivalent resistance R eq of the circuit and the total power P dissipated in it when the switch S is closed. Solution: 12V = 4Ω (a) I 1 = 4Ω + 2Ω = 2A , V 1 = (4Ω)(2A) = 8V . R 3 S (b) I 1 = 1 12V 2Ω + 2Ω = 1 . 5A , V 1 = (4Ω)(1 . 5A) = 6V . 2 „ 1 « − 1 = 4Ω 4Ω + 1 R 1 (c) R eq = + 2Ω = 4Ω , 4Ω = 2Ω R 2 P = (12V) 2 = 36W . 4Ω 12V 1/5/2019 [tsl378 – 13/60]

  4. Unit Exam II: Problem #3 (Spring ’08) Consider the electric circuit shown. Find the currents I 1 , I 2 , and I 3 (a) with the switch S open, (b) with the switch S closed. I 1 8V 2Ω 2Ω I 2 12V 2Ω S I 3 6V 1/5/2019 [tsl379 – 14/60]

  5. Unit Exam II: Problem #3 (Spring ’08) Consider the electric circuit shown. Find the currents I 1 , I 2 , and I 3 (a) with the switch S open, (b) with the switch S closed. I 1 8V 2Ω Solution: 2Ω (a) I 1 = 8V − 12V = − 1A , 4Ω I 2 12V I 2 = − I 1 = +1A . I 3 = 0 . (b) I 1 = 8V − 12V 2Ω = − 1A , S 4Ω I 3 = 6V − 12V = − 3A . I 2Ω 3 6V I 2 = − I 1 − I 3 = +4A . 1/5/2019 [tsl379 – 14/60]

  6. Unit Exam II: Problem #1 (Spring ’09) Both capacitor circuits are at equilibrium. (a) In the circuit on the left, the voltage across capacitor 1 is V 1 = 8V . Find the charge Q 1 on capacitor 1, the charge Q 2 on capacitor 2, and the voltage V 2 across capacitor 2. Find the emf E A supplied by the battery. (b) In the circuit on the right, the charge on capacitor 3 is Q 3 = 6 µ C . Find the voltage V 3 across capacitor 3, the voltage V 4 across capacitor 4, and the charge Q 4 on capacitor 4. Find the emf E B supplied by the battery. µ µ µ C = 1 F C = 2 F C = 3 F 2 1 3 µ F C = 4 4 ε ε B A 1/5/2019 [tsl392 – 15/60]

  7. Unit Exam II: Problem #1 (Spring ’09) Both capacitor circuits are at equilibrium. (a) In the circuit on the left, the voltage across capacitor 1 is V 1 = 8V . Find the charge Q 1 on capacitor 1, the charge Q 2 on capacitor 2, and the voltage V 2 across capacitor 2. Find the emf E A supplied by the battery. (b) In the circuit on the right, the charge on capacitor 3 is Q 3 = 6 µ C . Find the voltage V 3 across capacitor 3, the voltage V 4 across capacitor 4, and the charge Q 4 on capacitor 4. Find the emf E B supplied by the battery. µ µ µ C = 1 F C = 2 F C = 3 F 2 1 3 µ F C = 4 4 ε ε Solution: B A (a) Q 1 = (1 µ F)(8V) = 8 µ C , Q 2 = Q 1 = 8 µ C , V 2 = 8 µ C 2 µ F = 4V , E A = 8V + 4V = 12V . (b) V 3 = 6 µ C 3 µ F = 2V , V 4 = V 3 = 2V , Q 4 = (2V)(4 µ F) = 8 µ C , E B = V 3 = V 4 = 2V . 1/5/2019 [tsl392 – 15/60]

  8. Unit Exam II: Problem #2 (Spring ’09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq . (b) Find the power P supplied by the battery. 2Ω (c) Find the current I 4 through the 4Ω -resistor. 1Ω 1Ω (d) Find the voltage V 2 across the 2Ω -resistor. 4Ω 3Ω 3Ω 24V 1/5/2019 [tsl393 – 16/60]

  9. Unit Exam II: Problem #2 (Spring ’09) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq . (b) Find the power P supplied by the battery. 2Ω (c) Find the current I 4 through the 4Ω -resistor. 1Ω 1Ω (d) Find the voltage V 2 across the 2Ω -resistor. Solution: 4Ω 3Ω 3Ω (a) R eq = 8Ω . (b) P = (24V) 2 = 72W . 8Ω 24V (c) I 4 = 1 24V 8Ω = 1 . 5A . 2 (d) V 2 = (1 . 5A)(2Ω) = 3V . 1/5/2019 [tsl393 – 16/60]

  10. Unit Exam II: Problem #3 (Spring ’09) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 , and I 4 . 1Ω 1Ω I I 1 4 1V 3V 5V I I 2V 2 3 1Ω 1Ω 1/5/2019 [tsl394 – 17/60]

  11. Unit Exam II: Problem #3 (Spring ’09) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 , and I 4 . 1Ω 1Ω I I 1 4 1V Solution: 3V Use loops along quadrants in assumed current directions. Start at center. 5V +3V − I 1 (1Ω) − 1V = 0 ⇒ I 1 = 2A . I I 2V 2 3 +3V − I 2 (1Ω) + 2V = 0 ⇒ I 2 = 5A . − 2V − I 3 (1Ω) + 5V = 0 ⇒ I 3 = 3A . 1Ω 1Ω +1V − I 4 (1Ω) + 5V = 0 ⇒ I 4 = 6A . 1/5/2019 [tsl394 – 17/60]

  12. Unit Exam II: Problem #1 (Spring ’11) Both capacitor circuits are at equilibrium. (a) Find the charge Q 1 on capacitor 1. (b) Find the voltage V 3 across capacitor 3. (c) Find the charge Q 2 on capacitor 2. (d) Find the energy U 4 stored on capacitor 4. C C 1 = 1pF = 4pF 4 C = 2pF 24V 2 24V C 3 = 3pF 1/5/2019 [tsl404 – 18/60]

  13. Unit Exam II: Problem #1 (Spring ’11) Both capacitor circuits are at equilibrium. (a) Find the charge Q 1 on capacitor 1. (b) Find the voltage V 3 across capacitor 3. (c) Find the charge Q 2 on capacitor 2. (d) Find the energy U 4 stored on capacitor 4. C C 1 = 1pF = 4pF 4 C = 2pF 24V 2 24V Solution: C 3 = 3pF „ 1 « − 1 + 1 (a) C 13 = = 0 . 75pF , Q 1 = Q 3 = Q 13 = (24V)(0 . 75pF) = 18pC . C 1 C 3 (b) V 3 = Q 3 = 18pC 3pF = 6V . C 3 (c) Q 2 = (24V)(2pF) = 48pC . (d) U 4 = 1 4 = 1 2 (4pF)(24V) 2 = 1152pJ . 2 C 4 V 2 1/5/2019 [tsl404 – 18/60]

  14. Unit Exam II: Problem #2 (Spring ’11) Consider the resistor circuit shown. (a) Find the current I L on the left. (b) Find the current I R on the right. (c) Find the equivalent resistance R eq of all four resistors. (d) Find the power P 2 dissipated in resistor 2. = 1Ω = 4Ω R 1 R 4 I I 24V L R = 2Ω R 2 = 3Ω R 3 1/5/2019 [tsl405 – 19/60]

  15. Unit Exam II: Problem #2 (Spring ’11) Consider the resistor circuit shown. (a) Find the current I L on the left. (b) Find the current I R on the right. (c) Find the equivalent resistance R eq of all four resistors. (d) Find the power P 2 dissipated in resistor 2. = 1Ω = 4Ω R 1 R 4 Solution: I I 24V L R 24V (a) I L = 1Ω + 3Ω = 6A . = 2Ω R 2 (b) I R = 24V 4Ω = 6A . = 3Ω R 3 « − 1 „ 1Ω + 3Ω + 1 1 2Ω + 1 (c) R eq = = 1Ω . 4Ω (d) P 2 = (24V) 2 = 288W . 2Ω 1/5/2019 [tsl405 – 19/60]

  16. Unit Exam II: Problem #3 (Spring ’11) Consider the electric circuit shown. (a) Find the current I 1 . (b) Find the current I 2 . (c) Find the current I 3 . (d) Find the potential difference V a − V b . a 5Ω 3V I 10Ω 12V 2 I I 6V 1 3 b 1/5/2019 [tsl406 – 20/60]

  17. Unit Exam II: Problem #3 (Spring ’11) Consider the electric circuit shown. (a) Find the current I 1 . (b) Find the current I 2 . (c) Find the current I 3 . (d) Find the potential difference V a − V b . a 5Ω 3V I 10Ω 12V 2 I I 6V 1 3 b Solution: ⇒ I 1 = 15V (a) 12V + 3V − I 1 (10Ω) = 0 10Ω = 1 . 5A . ⇒ I 1 = 6V (b) − 6V + 12V − I 2 (5Ω) = 0 5Ω = 1 . 2A . (c) I 3 = I 1 + I 2 = 2 . 7A . (d) V a − V b = − 6V + 12V = 6V . 1/5/2019 [tsl406 – 20/60]

  18. Unit Exam II: Problem #1 (Spring ’12) Find the equivalent capacitances C eq of the two capacitor circuits. µ 3 n F 2 F 3 n F 3 n F µ 2 F 3 n F µ µ 2 F 2 F 1/5/2019 [tsl427 – 21/60]

  19. Unit Exam II: Problem #1 (Spring ’12) Find the equivalent capacitances C eq of the two capacitor circuits. µ 3 n F 2 F 3 n F 3 n F µ 2 F 3 n F µ µ 2 F 2 F Solution: „ 1 « − 1 1 1 • C eq = 3nF + 3nF + 3nF + = 4nF . 3nF « − 1 „ 1 1 1 = 4 • C eq = 2 µ F + 2 µ F + 2 µ F + 5 µ F . 2 µ F 1/5/2019 [tsl427 – 21/60]

  20. Unit Exam II: Problem #2 (Spring ’12) Consider a parallel-plate capacitor of capacitance C = 6 pF with plates separated a distance d = 1 mm and a potential difference V = V + − V − = 3 V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. −Q +Q (d) Find the area A of each plate. E V V − + d 1/5/2019 [tsl428 – 22/60]

  21. Unit Exam II: Problem #2 (Spring ’12) Consider a parallel-plate capacitor of capacitance C = 6 pF with plates separated a distance d = 1 mm and a potential difference V = V + − V − = 3 V between them. (a) Find the magnitude E of the electric field between the plates. (b) Find the amount Q of charge on each plate. (c) Find the energy U stored on the capacitor. −Q +Q (d) Find the area A of each plate. E Solution: (a) E = V 3V V V − + d = 1mm = 3000V / m . d (b) Q = CV = (6pF)(3V) = 18pC . (c) U = 1 2 QV = 0 . 5(18pC)(3V) = 27pJ . (d) A = Cd (6pF)(1mm) 8 . 85 × 10 − 12 C 2 N − 1 m − 2 = 6 . 78 × 10 − 4 m 2 . = ǫ 0 1/5/2019 [tsl428 – 22/60]

  22. Unit Exam II: Problem #3 (Spring ’12) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 , and I 4 I 2Ω 3 I 2 I 2Ω 12V 1 4Ω I 4 1/5/2019 [tsl429 – 23/60]

  23. Unit Exam II: Problem #3 (Spring ’12) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 , and I 4 I 2Ω 3 I 2 I 2Ω 12V 1 4Ω I 4 Solution: 12V • I 1 = 2Ω + 4Ω = 2A . • I 2 = 12V 2Ω = 6A . • I 3 = I 4 = I 1 + I 2 = 8A . 1/5/2019 [tsl429 – 23/60]

  24. Unit Exam II: Problem #4 (Spring ’12) Consider the electric circuit shown. Find the currents I 1 , I 2 , and I 3 8Ω I 1 12V I 3 6V I 3V 2 4Ω 1/5/2019 [tsl430 – 24/60]

  25. Unit Exam II: Problem #4 (Spring ’12) Consider the electric circuit shown. Find the currents I 1 , I 2 , and I 3 8Ω I 1 12V I 3 6V I 3V Solution: 2 4Ω ⇒ I 1 = 9 • 12V + 6V − (8Ω) I 1 = 0 4 A = 2 . 25A . ⇒ I 2 = 3 • 6V − 3V − (4Ω) I 2 = 0 4 A = 0 . 75A . • I 3 = I 1 + I 2 = 3 . 00A . 1/5/2019 [tsl430 – 24/60]

  26. Unit Exam II: Problem #1 (Spring ’13) Consider the capacitor circuit shown at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the total energy U stored in the four capacitors. (c) Find the voltage V ∗ across the capacitor marked by an asterisk. 5nF 6nF * 20V 4nF 5nF 1nF 10V 9nF 9nF 6nF * 1/5/2019 [tsl456 – 25/60]

  27. Unit Exam II: Problem #1 (Spring ’13) Consider the capacitor circuit shown at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the total energy U stored in the four capacitors. (c) Find the voltage V ∗ across the capacitor marked by an asterisk. 5nF 6nF * 20V 4nF 5nF 1nF 10V 9nF 9nF 6nF * Solution: « − 1 « − 1 „ 1 1 1 „ 1 1 1 C eq = 5nF + 1nF + 6nF + C eq = 4nF + 5nF + 9nF + 6nF 9nF = 2nF = 3nF U = 1 U = 1 2 (2nF)(10V) 2 = 100nJ 2 (3nF)(20V) 2 = 600nJ V ∗ = 10 V ∗ = 20 3 V = 3 . 33V 3 V = 6 . 67V 1/5/2019 [tsl456 – 25/60]

  28. Unit Exam II: Problem #2 (Spring ’13) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq . (b) Find the current I flowing through the battery. (c) Find the voltage V ∗ across the resistor marked by an asterisk. 6Ω 1Ω * 8Ω 8Ω 20V * 6Ω 20V 3Ω 3Ω 1Ω 1/5/2019 [tsl457 – 26/60]

  29. Unit Exam II: Problem #2 (Spring ’13) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq . (b) Find the current I flowing through the battery. (c) Find the voltage V ∗ across the resistor marked by an asterisk. 6Ω 1Ω * 8Ω 8Ω 20V * 6Ω 20V 3Ω 3Ω 1Ω Solution: „ 1 „ 1 « − 1 « − 1 8Ω + 1 6Ω + 1 R eq = + 3Ω + 3Ω = 10Ω R eq = + 1Ω + 1Ω = 5Ω 8Ω 6Ω I = 20V I = 20V 10Ω = 2A 5Ω = 4A V ∗ = (1A)(8Ω) = 8V V ∗ = (2A)(6Ω) = 12V 1/5/2019 [tsl457 – 26/60]

  30. Unit Exam II: Problem #3 (Spring ’13) Consider the RC circuit shown. The switch has been closed for a long time. (a) Find the current I B flowing through the battery. (b) Find the voltage V C across the capacitor. (c) Find the charge Q on the capacitor. (d) Find the current I 3 flowing through the 3Ω -resistor right after the switch has been opened. 4Ω 4Ω 10nF 2Ω 1Ω 12V 12V 10nF 3Ω 3Ω S S 1/5/2019 [tsl458 – 27/60]

  31. Unit Exam II: Problem #3 (Spring ’13) Consider the RC circuit shown. The switch has been closed for a long time. (a) Find the current I B flowing through the battery. (b) Find the voltage V C across the capacitor. (c) Find the charge Q on the capacitor. (d) Find the current I 3 flowing through the 3Ω -resistor right after the switch has been opened. 4Ω 4Ω 10nF 2Ω 1Ω 12V 12V 10nF 3Ω 3Ω S S Solution: 12V 12V I B = 2Ω + 4Ω = 2A I B = 3Ω + 1Ω + 4Ω = 1 . 5A V C = (2A)(2Ω) = 4V V C = (1 . 5A)(3Ω + 1Ω) = 6V Q = (4V)(10nF) = 40nC Q = (6V)(10nF) = 60nC 4V 6V I 3 = 2Ω + 3Ω = 0 . 8A I 3 = 3Ω + 1Ω = 1 . 5A 1/5/2019 [tsl458 – 27/60]

  32. Unit Exam II: Problem #1 (Spring ’14) Both capacitor circuits, charged up by batteries as shown, are now at equilibrium. The charge on capacitor C 1 = 6 pF [8pF] is Q 1 = 18 pC [16pF] and charge on capacitor C 4 = 8 pF [4pf] is Q 4 = 16 pC [12pF]. (a) Find the voltage V 2 across capacitor C 2 = 4 pF . (b) Find the emf E A supplied by the battery. (c) Find the charge Q 3 on capacitor C 3 = 3 pF . (d) Find the emf E B supplied by the battery. C 1 ε ε B C 3 C 4 A C 2 1/5/2019 [tsl472 – 28/60]

  33. Unit Exam II: Problem #1 (Spring ’14) Both capacitor circuits, charged up by batteries as shown, are now at equilibrium. The charge on capacitor C 1 = 6 pF [8pF] is Q 1 = 18 pC [16pF] and charge on capacitor C 4 = 8 pF [4pf] is Q 4 = 16 pC [12pF]. (a) Find the voltage V 2 across capacitor C 2 = 4 pF . (b) Find the emf E A supplied by the battery. (c) Find the charge Q 3 on capacitor C 3 = 3 pF . (d) Find the emf E B supplied by the battery. C 1 ε ε B C 3 C 4 A C 2 Solution: V 2 = Q 2 (a) Q 2 = Q 1 = 18pC , [16pC] , = 4 . 5V [4V] . C 2 (b) E A = V 1 + V 2 = 3V + 4 . 5V = 7 . 5V [2V + 4V = 6V] . (c) V 3 = V 4 = Q 4 = 2V [3V] , Q 3 = V 3 C 3 = 6pC [9pC] . C 4 (d) E B = V 3 = V 4 = 2V [3V] . 1/5/2019 [tsl472 – 28/60]

  34. Unit Exam II: Problem #2 (Spring ’14) Consider the resistor circuit shown with R 1 = 2Ω [ 3Ω ], R 2 = 3Ω [ 2Ω ], and R 3 = 1Ω . (a) Find the current I 2 through resistor R 2 . (b) Find the voltage V 3 across resitor R 3 . (c) Find the power P 1 dissipated in resistor R 1 . (d) Find the equivalent resistance R eq . R 12V 1 R 2 R 3 1/5/2019 [tsl473 – 29/60]

  35. Unit Exam II: Problem #2 (Spring ’14) Consider the resistor circuit shown with R 1 = 2Ω [ 3Ω ], R 2 = 3Ω [ 2Ω ], and R 3 = 1Ω . (a) Find the current I 2 through resistor R 2 . (b) Find the voltage V 3 across resitor R 3 . (c) Find the power P 1 dissipated in resistor R 1 . (d) Find the equivalent resistance R eq . R 12V 1 R 2 Solution: R 3 12V » 12V – (a) I 2 = 3Ω + 1Ω = 3A 2Ω + 1Ω = 4A . (b) V 3 = (3A)(1Ω) = 3V [(4A)(1Ω) = 4V] . » (12V) 2 (c) P 1 = (12V) 2 – = 72W = 48W . 2Ω 3Ω „ 1 "„ 1 # « − 1 « − 1 1 = 4 1 = 3 (d) R eq = 2Ω + 3 Ω 3Ω + 2 Ω . 3Ω + 1Ω 2Ω + 1Ω 1/5/2019 [tsl473 – 29/60]

  36. Unit Exam II: Problem #3 (Spring ’14) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 , I 4 when ... (a) only switch S A is closed, (a) only switch S C is closed, (b) only switch S B is closed, (b) only switch S B is closed, (c) switches S A and S B are closed. (c) switches S B and S C are closed. S B S A S C 6V 3V 2V 4V I I I I 3 1 2 4 5Ω 5Ω 5Ω 1/5/2019 [tsl474 – 30/60]

  37. Unit Exam II: Problem #3 (Spring ’14) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 , I 4 when ... (a) only switch S A is closed, (a) only switch S C is closed, (b) only switch S B is closed, (b) only switch S B is closed, (c) switches S A and S B are closed. (c) switches S B and S C are closed. S B S A S C 6V 3V 2V 4V I I I I 3 1 2 4 5Ω 5Ω 5Ω Solution: (a) I 1 = 0 . 6A , I 2 = − 0 . 6A , I 3 = 0 , I 4 = 0 . (a) I 1 = 0 , I 2 = 0 , I 3 = − 0 . 4A , I 4 = 0 . 4A . (b) I 1 = 0 , I 2 = 0 . 2A , I 3 = − 0 . 2A , I 4 = 0 . (b) I 1 = 0 , I 2 = 0 . 2A , I 3 = − 0 . 2A , I 4 = 0 . (c) I 1 = 0 . 6A , I 2 = − 0 . 4A , (c) I 1 = 0 , I 2 = 0 . 2A , I 3 = − 0 . 2A , I 4 = 0 . I 3 = − 0 . 6A , I 4 = 0 . 4A . 1/5/2019 [tsl474 – 30/60]

  38. Unit Exam II: Problem #1 (Fall ’14) Both capacitor circuits, charged up by batteries as shown, are now at equilibrium. Each of the six capacitors has a 2pF capacitance. (a) Find the equivalent capacitance of the circuit on the left. (b) Find the voltages V 1 , V 2 , V 3 across capacitors C 1 , C 2 , C 3 , respectively. (c) Find the equivalent capacitance of the circuit on the right. (d) Find the charges Q 4 , Q 5 , Q 6 on capacitors C 4 , C 5 , C 6 , respectively. C C 1 4 C C C 3 5 2 C 12V 6 12V 1/5/2019 [tsl482 – 31/60]

  39. Unit Exam II: Problem #1 (Fall ’14) Both capacitor circuits, charged up by batteries as shown, are now at equilibrium. Each of the six capacitors has a 2pF capacitance. (a) Find the equivalent capacitance of the circuit on the left. (b) Find the voltages V 1 , V 2 , V 3 across capacitors C 1 , C 2 , C 3 , respectively. (c) Find the equivalent capacitance of the circuit on the right. (d) Find the charges Q 4 , Q 5 , Q 6 on capacitors C 4 , C 5 , C 6 , respectively. C C 1 4 C C C 3 5 2 Solution: „ 1 « − 1 C 1 12V 6 12V (a) C eq = 2pF + 2pF + = 3pF . 2pF (b) V 1 = 12V , V 2 = V 3 = 6V « − 1 „ 1 1 = 4 (c) C eq = 2pF + 2pF + 3 pF . 2pF (d) Q 45 = Q 6 = C eq (12V) = 16pC ⇒ Q 4 = Q 5 = 8pC . 1/5/2019 [tsl482 – 31/60]

  40. Unit Exam II: Problem #2 (Fall ’14) Consider the resistor circuit shown with R 1 = 5Ω , R 2 = 1Ω , and R 3 = 3Ω . (a) Find the equivalent resistance R eq . (b) Find the currents I 1 , I 2 , I 3 through resistors R 1 , R 2 , R 3 , respectively. (c) Find the voltages V 1 , V 2 , V 3 across resistors R 1 , R 2 , R 3 , respectively. R 3 R 12V 1 R 2 1/5/2019 [tsl483 – 32/60]

  41. Unit Exam II: Problem #2 (Fall ’14) Consider the resistor circuit shown with R 1 = 5Ω , R 2 = 1Ω , and R 3 = 3Ω . (a) Find the equivalent resistance R eq . (b) Find the currents I 1 , I 2 , I 3 through resistors R 1 , R 2 , R 3 , respectively. (c) Find the voltages V 1 , V 2 , V 3 across resistors R 1 , R 2 , R 3 , respectively. R 3 R 12V 1 R 2 Solution: « − 1 „ 1Ω + 3Ω + 1 1 = 20 (a) R eq = 9 Ω = 2 . 22Ω . 5Ω (b) I 1 = 12V 12V 5Ω = 2 . 4A , I 2 = I 3 = 1Ω + 3Ω = 3A . (c) V 1 = R 1 I 1 = 12V , V 2 = R 2 I 2 = 3V , V 3 = R 3 I 3 = 9V . 1/5/2019 [tsl483 – 32/60]

  42. Unit Exam II: Problem #3 (Fall ’14) Consider the two-loop circuit shown. (a) Find the current I 1 . (b) Find the current I 2 . (c) Find the potential difference V a − V b . 4V 2V a 3Ω 5Ω 6V b I I 2 1 1/5/2019 [tsl484 – 33/60]

  43. Unit Exam II: Problem #3 (Fall ’14) Consider the two-loop circuit shown. (a) Find the current I 1 . (b) Find the current I 2 . (c) Find the potential difference V a − V b . 4V 2V a 3Ω 5Ω 6V b I I 2 1 Solution: (a) I 1 = 6V − 4V = 0 . 4A . 5Ω (b) I 2 = 6V + 2V = 2 . 67A . 3Ω (c) V a − V b = 6V + 2V = 8V . 1/5/2019 [tsl484 – 33/60]

  44. Unit Exam II: Problem #1 (Spring ’15) Both capacitor circuits are at equilibrium. (a) Find the charge Q 1 on capacitor 1. (b) Find the energy U 3 stored on capacitor 3. (c) Find the charge Q 2 on capacitor 2. 24V (d) Find the voltage V 4 across capacitor 4. C = 3pF 3 C 2 = 2pF C = 1pF 1 C = 4pF 4 24V 1/5/2019 [tsl491 – 34/60]

  45. Unit Exam II: Problem #1 (Spring ’15) Both capacitor circuits are at equilibrium. (a) Find the charge Q 1 on capacitor 1. (b) Find the energy U 3 stored on capacitor 3. (c) Find the charge Q 2 on capacitor 2. 24V (d) Find the voltage V 4 across capacitor 4. C = 3pF 3 C 2 = 2pF C = 1pF 1 Solution: C = 4pF 4 (a) Q 1 = C 1 V 1 = (1pF)(24V) = 24pC . 24V (b) U 3 = 1 3 = 1 2 (3pF)(24V) 2 = 864pJ . 2 C 3 V 2 „ 1 « − 1 + 1 = 4 (c) C 24 = 3 pF , C 2 C 4 „ 4 « Q 2 = Q 4 = Q 24 = C 24 V 24 = 3 pF (24V) = 32pC . (d) V 4 = Q 4 = 32pC 4pF = 8V . C 4 1/5/2019 [tsl491 – 34/60]

  46. Unit Exam II: Problem #2 (Spring ’15) In the two resistor circuits shown find the equivalent resistances R 123 (left) and R 456 (right). Then find the currents I 1 , I 2 , I 3 through the individual resistors on the left. and the currents I 4 , I 5 , I 6 through the individual resistors on the right. 2Ω 2Ω R = R = 4 1 2Ω R = 2Ω 2Ω 2Ω R = R = R = 2 3 5 6 14V 14V 1/5/2019 [tsl492 – 35/60]

  47. Unit Exam II: Problem #2 (Spring ’15) In the two resistor circuits shown find the equivalent resistances R 123 (left) and R 456 (right). Then find the currents I 1 , I 2 , I 3 through the individual resistors on the left. and the currents I 4 , I 5 , I 6 through the individual resistors on the right. 2Ω 2Ω R = R = 4 1 2Ω R = 2Ω 2Ω 2Ω R = R = R = 2 3 5 6 14V 14V Solution: „ 1 « − 1 2Ω + 1 = 4 • R 23 = 2Ω + 2Ω = 4Ω , R 123 = 3 Ω 4Ω „ 1 « − 1 2Ω + 1 • R 45 = = 1Ω , R 456 = R 45 + R 6 = 3Ω 2Ω • I 1 = 14V I 2 = I 3 = 14V 2Ω = 7A , 4Ω = 3 . 5A • I 6 = I 45 = 14V I 4 = I 5 = 1 3Ω = 4 . 67A , 2 I 6 = 2 . 33A 1/5/2019 [tsl492 – 35/60]

  48. Unit Exam II: Problem #3 (Spring ’15) In the circuit shown find the currents I 1 , I 2 , and the potential difference V b − V a (a) if the switch S is open, (b) if the switch S is closed. 8V 4V a S 3Ω 2Ω b I I 2 1 1/5/2019 [tsl493 – 36/60]

  49. Unit Exam II: Problem #3 (Spring ’15) In the circuit shown find the currents I 1 , I 2 , and the potential difference V b − V a (a) if the switch S is open, (b) if the switch S is closed. 8V 4V a S 3Ω 2Ω b I I 2 1 Solution: (a) I 1 = I 2 = 12V 5Ω = 2 . 4A V b − V a = 8V − (2 . 4A)(2Ω) = − 4V + (2 . 4A)(3Ω) = 3 . 2V . (b) I 1 = 8V I 2 = 4V 2Ω = 4A , 3Ω = 1 . 33A , V b − V a = 0 . 1/5/2019 [tsl493 – 36/60]

  50. Unit Exam II: Problem #1 (Fall ’15) Consider the capacitor circuit shown at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the total energy U stored in the three capacitors. (c) Find the voltage V ∗ across the capacitor marked by an asterisk. (d) Find the voltage V 1 across the 1nF-capacitor. 6V 8V 4nF * 2nF 3nF 3nF * 1nF 1nF 1/5/2019 [tsl515 – 37/60]

  51. Unit Exam II: Problem #1 (Fall ’15) Consider the capacitor circuit shown at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the total energy U stored in the three capacitors. (c) Find the voltage V ∗ across the capacitor marked by an asterisk. (d) Find the voltage V 1 across the 1nF-capacitor. 6V 8V 4nF * 2nF 3nF 3nF * 1nF 1nF Solution: « − 1 « − 1 „ 1 1 „ 1 1 (a) C eq = 1nF + 2nF + = 1 . 5nF (a) C eq = 3nF + 1nF + = 2nF 3nF 4nF (b) U = 1 (b) U = 1 2 (1 . 5nF)(6V) 2 = 27nJ 2 (2nF)(8V) 2 = 64nJ (c) V ∗ = 1 (c) V ∗ = 1 26V = 3V 2 8V = 4V (d) V 1 = 6V − 3V = 3V (d) V 1 = 8V − 4V = 4V 1/5/2019 [tsl515 – 37/60]

  52. Unit Exam II: Problem #2 (Fall ’15) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq . (b) Find the currents I 1 and I 2 . (c) Find the power P supplied by the battery. I 6V I 8V 1 3Ω 1 4Ω 2Ω 3Ω 2Ω 4Ω I I 2 2 1/5/2019 [tsl516 – 38/60]

  53. Unit Exam II: Problem #2 (Fall ’15) Consider the resistor circuit shown. (a) Find the equivalent resistance R eq . (b) Find the currents I 1 and I 2 . (c) Find the power P supplied by the battery. I 6V I 8V 1 3Ω 1 4Ω 2Ω 3Ω 2Ω 4Ω I I 2 2 Solution: „ 1 „ 1 « − 1 « − 1 4Ω + 1 2Ω + 1 (a) R eq = + 3Ω = 5Ω (a) R eq = + 3Ω = 4Ω 4Ω 2Ω (b) I 1 = 6V I 2 = 1 (b) I 1 = 8V I 2 = 1 5Ω = 1 . 2A , 2 I 1 = 0 . 6A 4Ω = 2A , 2 I 1 = 1A (c) P = (1 . 2A)(6V) = 7 . 2W (c) P = (2A)(8V) = 16W 1/5/2019 [tsl516 – 38/60]

  54. Unit Exam II: Problem #3 (Fall ’15) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 . 2Ω 3Ω 2Ω 3Ω 3V 3V 12V 3V 12V 3V I I I I I I 2 3 2 3 1 1 1/5/2019 [tsl517 – 39/60]

  55. Unit Exam II: Problem #3 (Fall ’15) Consider the electric circuit shown. Find the currents I 1 , I 2 , I 3 . 2Ω 3Ω 2Ω 3Ω 3V 3V 12V 3V 12V 3V I I I I I I 2 3 2 3 1 1 Solution: 12V − I 2 (2Ω) − 3V = 0 12V − I 2 (2Ω) + 3V = 0 ⇒ I 2 = 9V ⇒ I 2 = 15V 2Ω = 4 . 5A 2Ω = 7 . 5A . 12V − I 3 (3Ω) + 3V = 0 12V − I 3 (3Ω) − 3V = 0 ⇒ I 3 = 15V ⇒ I 3 = 9V 3Ω = 5A . 3Ω = 3A . I 1 = I 2 + I 3 = 9 . 5A I 1 = I 2 + I 3 = 10 . 5A 1/5/2019 [tsl517 – 39/60]

  56. Unit Exam II: Problem #1 (Spring ’16) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the total energy U stored in the three capacitors. (c) Find the charge Q 6 on the capacitor on the left. (d) Find the the voltages V 2 and V 4 across the two capacitor on the right. 2 µ F 6 µ F 4 µ F 8V 1/5/2019 [tsl529 – 40/60]

  57. Unit Exam II: Problem #1 (Spring ’16) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the total energy U stored in the three capacitors. (c) Find the charge Q 6 on the capacitor on the left. (d) Find the the voltages V 2 and V 4 across the two capacitor on the right. Solution: 2 µ F 6 µ F « − 1 „ 1 1 4 µ F (a) C eq = 2 µ F + 4 µ F + = 3 µ F . 6 µ F 8V (b) U = 1 2 (3 µ F)(8V) 2 = 96 µ J . (c) Q 6 = (8V)(3 µ F) = 24 µ C . (d) V 2 = V 4 = 1 2 (8V) = 4V . 1/5/2019 [tsl529 – 40/60]

  58. Unit Exam II: Problem #2 (Spring ’16) Consider the electrical circuit shown. (a) Find the current I 1 when the switch S is open. (b) Find the currents I 1 and I 2 when the switch S is closed. 2Ω 3Ω 4V 5Ω S I 2 4Ω I 1 6V 1/5/2019 [tsl530 – 41/60]

  59. Unit Exam II: Problem #2 (Spring ’16) Consider the electrical circuit shown. (a) Find the current I 1 when the switch S is open. (b) Find the currents I 1 and I 2 when the switch S is closed. 2Ω 3Ω 4V 5Ω S Solution: I 2 6V − 4V (a) I 1 = 4Ω + 5Ω + 3Ω + 2Ω = 0 . 143A . 4Ω I 1 6V 6V 4V (b) I 1 = 4Ω + 5Ω = 0 . 667A , I 2 = 3Ω + 2Ω = 0 . 8A . 1/5/2019 [tsl530 – 41/60]

  60. Unit Exam II: Problem #3 (Spring ’15) This RC circuit has been running for a long time with the switch open. (a) Find the current I while the switch is still open. (b) Find the current I right after the switch has been closed. (c) Find the current I a long time later. (d) Find the charge Q on the capacitor also a long time later. 2Ω 4Ω S 12V 7nF I 1/5/2019 [tsl531 – 42/60]

  61. Unit Exam II: Problem #3 (Spring ’15) This RC circuit has been running for a long time with the switch open. (a) Find the current I while the switch is still open. (b) Find the current I right after the switch has been closed. (c) Find the current I a long time later. (d) Find the charge Q on the capacitor also a long time later. 2Ω Solution: 4Ω S 12V 12V (a) I = 2Ω + 4Ω = 2A . 7nF I (b) I = 12V 2Ω = 6A . 12V (c) I = 2Ω + 4Ω = 2A . (d) Q = (8V)(7nF) = 56nC . 1/5/2019 [tsl531 – 42/60]

  62. Unit Exam II: Problem #1 (Fall ’16) The capacitors (initially discharged) have been connected to the battery. The circuit is now at equilibrium. Find ... (a) the voltage V 2 across capacitor C 2 , (a) the voltage V 4 across capacitor C 4 , (b) the energy U 5 on capacitor C 5 , (b) the energy U 7 on capacitor C 7 , (c) the charge Q 3 on capacitor C 3 , (c) the charge Q 6 on capacitor C 6 , (d) the equivalent capacitance C eq . (d) the equivalent capacitance C eq . C 4 = 4 µ F 12V C 3 = 3 µ F C 5 = 5 µ F µ C 6 = 6 µ F C = 6 F 6 C 7 = 7 µ F C 3 = 3 µ F C 2 = 2 µ F 18V 1/5/2019 [tsl538 – 43/60]

  63. Unit Exam II: Problem #1 (Fall ’16) The capacitors (initially discharged) have been connected to the battery. The circuit is now at equilibrium. Find ... (a) the voltage V 2 across capacitor C 2 , (a) the voltage V 4 across capacitor C 4 , (b) the energy U 5 on capacitor C 5 , (b) the energy U 7 on capacitor C 7 , (c) the charge Q 3 on capacitor C 3 , (c) the charge Q 6 on capacitor C 6 , (d) the equivalent capacitance C eq . (d) the equivalent capacitance C eq . Solution: (a) V 2 = 12V . (a) V 4 = 18V . (b) U 5 = 1 (b) U 7 = 1 2 (5 µ F)(12V) 2 = 360 µ J . 2 (7 µ F)(18V) 2 = 1134 µ J . (c) C 36 = 2 µ F (c) C 36 = 2 µ F ⇒ Q 3 = Q 36 = (12V)(2 µ F) = 24 µ C . ⇒ Q 6 = Q 36 = (18V)(2 µ F) = 36 µ C . (d) C eq = C 5 + C 36 + C 2 = 9 µ F . (d) C eq = C 4 + C 36 + C 7 = 13 µ F . 1/5/2019 [tsl538 – 43/60]

  64. Unit Exam II: Problem #2 (Fall ’16) This resistor circuit is in a state of steady currents. Find ... (a) the voltage V 2 across resistor R 2 , (a) the voltage V 3 across resistor R 3 , (b) the power P 4 dissipated in resistor R 4 , (b) the power P 6 dissipated in resistor R 6 , (c) the current I 3 flowing through resistor R 3 (c) the current I 4 flowing through resistor R 4 , (d) the equivalent resistance R eq . (d) the equivalent resistance R eq . 3Ω 18V R = 3 4Ω R = 2Ω R = 4 2 3Ω 4Ω R = R = 3 4 1Ω 6Ω R = R = 1 6 2Ω R = 12V 2 1/5/2019 [tsl539 – 44/60]

  65. Unit Exam II: Problem #2 (Fall ’16) This resistor circuit is in a state of steady currents. Find ... (a) the voltage V 2 across resistor R 2 , (a) the voltage V 3 across resistor R 3 , (b) the power P 4 dissipated in resistor R 4 , (b) the power P 6 dissipated in resistor R 6 , (c) the current I 3 flowing through resistor R 3 (c) the current I 4 flowing through resistor R 4 , (d) the equivalent resistance R eq . (d) the equivalent resistance R eq . Solution: (a) V 2 = 18V . (a) V 3 = 12V (b) P 4 = 18V 2 (b) P 6 = 12V 2 = 81W . = 24W . 4Ω 6Ω 18V 12V (c) I 3 = 3Ω + 1Ω = 4 . 5A . (c) I 4 = 2Ω + 4Ω = 2A . „ 1 „ 1 « − 1 « − 1 1Ω + 3Ω + 1 1 2Ω + 4Ω + 1 1 (d) R eq = 4Ω + = 1Ω . (d) R eq = 3Ω + = 1 . 5Ω 2Ω 6Ω 1/5/2019 [tsl539 – 44/60]

  66. Unit Exam II: Problem #3 (Fall ’16) This two-loop resistor circuit is in a state of steady currents. Find ... (a) the current I 1 , (b) the current I 2 , (c) the potential difference V a − V b . 7V 11V 7V 11V a a I I I I 5V 5V 1 2 1 2 b b 6Ω 8Ω 8Ω 6Ω 1/5/2019 [tsl540 – 45/60]

  67. Unit Exam II: Problem #3 (Fall ’16) This two-loop resistor circuit is in a state of steady currents. Find ... (a) the current I 1 , (b) the current I 2 , (c) the potential difference V a − V b . 7V 11V 7V 11V a a I I I I 5V 5V 1 2 1 2 b b 6Ω 8Ω 8Ω 6Ω Solution: (a) I 1 = 5V + 7V (a) I 1 = 7V − 5V = +1 . 5A . = +0 . 333A . 8Ω 6Ω (b) I 2 = 5V + 11V (b) I 2 = 5V + 11V = +2 . 67A . = +2A . 6Ω 8Ω (c) V a − V b = − 7V + 11V = +4V . (c) V a − V b = 7V + 11V = +18V . 1/5/2019 [tsl540 – 45/60]

  68. Unit Exam II: Problem #1 (Spring ’17) The capacitors (initially discharged) have been connected to the battery. The circuit is now at equilibrium. Find ... (a) the charge Q 4 on the 4pF-capacitor, (a) the charge Q 3 on the 3pF-capacitor, (b) the energy U 7 on the 7pF-capacitor, (b) the energy U 5 on the 5pF-capacitor, (c) the voltage V 10 across the upper (c) the voltage V 8 across the lower 10pF-capacitor, 8pF-capacitor, (d) the equivalent capacitance C eq . (d) the equivalent capacitance C eq . 4pF 10pF 8pF 9V 7pF 5pF 6V 10pF 8pF 3pF 1/5/2019 [tsl548 – 46/60]

  69. Unit Exam II: Problem #1 (Spring ’17) The capacitors (initially discharged) have been connected to the battery. The circuit is now at equilibrium. Find ... (a) the charge Q 4 on the 4pF-capacitor, (a) the charge Q 3 on the 3pF-capacitor, (b) the energy U 7 on the 7pF-capacitor, (b) the energy U 5 on the 5pF-capacitor, (c) the voltage V 10 across the upper (c) the voltage V 8 across the lower 10pF-capacitor, 8pF-capacitor, (d) the equivalent capacitance C eq . (d) the equivalent capacitance C eq . Solution: (a) Q 4 = (6V)(4pF) = 24pC . (a) Q 3 = (9V)(3pF) = 27pC . (b) U 7 = 1 (b) U 5 = 1 2 (7pF)(6V) 2 = 126pJ . 2 (5pF)(9V) 2 = 202 . 5pJ . (c) V 10 = 1 (c) V 8 = 1 2 6V = 3V . 2 9V = 4 . 5V . (d) C eq = 4pF + 7pF + 5pF = 16pF . (d) C eq = 3pF + 5pF + 4pF = 12pF . 1/5/2019 [tsl548 – 46/60]

  70. Unit Exam II: Problem #2 (Spring ’17) Consider this circuit with two terminals, four resistors, and one switch. (a) Find the equivalent resistance R (open) when the switch is open. eq (b) Find the equivalent resistance R (closed) when the switch is closed. eq 2Ω 1Ω 3Ω 1Ω S S 1Ω 2Ω 1Ω 3Ω 1/5/2019 [tsl549 – 47/60]

  71. Unit Exam II: Problem #2 (Spring ’17) Consider this circuit with two terminals, four resistors, and one switch. (a) Find the equivalent resistance R (open) when the switch is open. eq (b) Find the equivalent resistance R (closed) when the switch is closed. eq 2Ω 1Ω 3Ω 1Ω S S 1Ω 2Ω 1Ω 3Ω Solution: « − 1 « − 1 „ 1 1 = 3 „ 1 1 R (open) R (open) = 1Ω + 2Ω + 2 Ω . = 1Ω + 3Ω + = 2Ω . eq eq 1Ω + 2Ω 1Ω + 3Ω „ 1 „ 1 „ 1 „ 1 « − 1 « − 1 « − 1 « − 1 1Ω + 1 1Ω + 1 1Ω + 1 1Ω + 1 R (closed) R (closed) = + = + eq eq 2Ω 2Ω 3Ω 3Ω = 4 = 3 3 Ω . 2 Ω . 1/5/2019 [tsl549 – 47/60]

  72. Unit Exam II: Problem #3 (Spring ’17) Consider this circuit with two batteries, two resistors, and one switch. (a) Find the current I when the switch is open. (b) Find the current I when the switch is closed. (c) Find the potential difference V a − V b when the switch is open. (d) Find the potential difference V a − V b when the switch is closed. 15V 16V 15V 12V a a S S I I b b 5Ω 6Ω 5Ω 2Ω 1/5/2019 [tsl550 – 48/60]

  73. Unit Exam II: Problem #3 (Spring ’17) Consider this circuit with two batteries, two resistors, and one switch. (a) Find the current I when the switch is open. (b) Find the current I when the switch is closed. (c) Find the potential difference V a − V b when the switch is open. (d) Find the potential difference V a − V b when the switch is closed. Solution: (a) I = 15V (a) I = 16V 5Ω = 3A . 2Ω = 8A . (b) I = 15V 5Ω + 12V (b) I = 16V 2Ω + 15V 6Ω = 3A + 2A = 5A . 5Ω = 8A + 3A = 11A . (c) V a − V b = 12V . (c) V a − V b = 15V . (d) V a − V b = 0 . (d) V a − V b = 0 . 1/5/2019 [tsl550 – 48/60]

  74. Unit Exam II: Problem #1 (Fall ’17) This circuit is at equilibrium. • Find the charge Q 7 on capacitor C 7 [ Q 5 on C 5 ]. • Find the energy U 5 on capacitor C 5 [ U 7 on C 7 ]. • Find the voltages V 2 , V 4 across capacitors C 2 , C 4 [ V 3 , V 6 across C 3 , C 6 ]. C 5 = 5 µ F µ F C = 2 C 3 = 3 µ F 2 24V C 4 = 4 µ F = 6 µ F C 6 C 7 = 7 µ F 1/5/2019 [tsl557 – 49/60]

  75. Unit Exam II: Problem #1 (Fall ’17) This circuit is at equilibrium. • Find the charge Q 7 on capacitor C 7 [ Q 5 on C 5 ]. • Find the energy U 5 on capacitor C 5 [ U 7 on C 7 ]. • Find the voltages V 2 , V 4 across capacitors C 2 , C 4 [ V 3 , V 6 across C 3 , C 6 ]. C 5 = 5 µ F µ F C = 2 C 3 = 3 µ F 2 24V C 4 = 4 µ F = 6 µ F C 6 C 7 = 7 µ F Solution: • Q 7 = (24V)(7 µ F) = 168 µ C [ Q 5 = (24V)(5 µ F) = 120 µ C] • » – • U 5 = 1 U 7 = 1 2 (5 µ F)(24V) 2 = 1440 µ J 2 (7 µ F)(24V) 2 = 2016 µ J • V 2 + V 4 = 24V , V 2 C 2 = V 4 C 4 ⇒ V 2 = 16V , V 4 = 8V [ V 3 + V 6 = 24V , V 3 C 3 = V 6 C 6 ⇒ V 3 = 16V , V 6 = 8V] 1/5/2019 [tsl557 – 49/60]

  76. Unit Exam II: Problem #2 (Fall ’17) Consider the resistor circuit on the left [right]. Find the currents I 1 , I 2 [ I 3 , I 4 ] and the potential difference V a − V b [ V c − V d ] (a) when the switch S w [ S y ] is open, (b) when the switch S w [ S y ] is closed 3Ω 5Ω 6Ω 4Ω b a d c S w S y 3V 6V 2V 5V I I I 3 I 1 2 4 1/5/2019 [tsl558 – 50/60]

  77. Unit Exam II: Problem #2 (Fall ’17) Consider the resistor circuit on the left [right]. Find the currents I 1 , I 2 [ I 3 , I 4 ] and the potential difference V a − V b [ V c − V d ] (a) when the switch S w [ S y ] is open, (b) when the switch S w [ S y ] is closed 3Ω 5Ω 6Ω 4Ω b a d c S w S y 3V 6V 2V 5V I I I 3 I 1 2 4 Solution: (a) I 1 = I 2 = 3V + 6V 5Ω + 3Ω = 1 . 125A , V a − V b = 9 V. » I 3 = I 4 = 2V + 5V – 6Ω + 4Ω = 0 . 7A , V c − V d = 7 V. (b) I 1 = 3V I 2 = 6V 5Ω = 0 . 6A , 3Ω = 2A , V a − V b = 9 V. » I 3 = 5V I 4 = 2V – 4Ω = 1 . 25A , 6Ω = 0 . 333A , V c − V d = 7 V. 1/5/2019 [tsl558 – 50/60]

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