RC circuits with DC sources A Circuit i (resistors, voltage - PowerPoint PPT Presentation

Plot of f ( t ) = A e − t /τ + B * At t = 0, f = A + B . * As t → ∞ , f → B . * The graph of f ( t ) lies between ( A + B ) and B . Note: If A > 0, A + B > B . If A < 0, A + B < B . * At t = 0, df � − 1 � = − A dt = A e − t /τ τ . τ If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive. * As t → ∞ , df dt → 0, i.e., f becomes constant (equal to B ). A + B B A + B B 0 5 τ t 0 5 τ t A < 0 A > 0 M. B. Patil, IIT Bombay

RL circuits with DC sources A Circuit i (resistors, voltage sources, v L current sources, CCVS, CCCS, VCVS, VCCS) B

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B M. B. Patil, IIT Bombay

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), V Th = constant . M. B. Patil, IIT Bombay

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), V Th = constant . * KVL: V Th = R Th i + L di dt . M. B. Patil, IIT Bombay

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), V Th = constant . * KVL: V Th = R Th i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L / R Th → i ( h ) = K exp( − t /τ ) . M. B. Patil, IIT Bombay

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), V Th = constant . * KVL: V Th = R Th i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L / R Th → i ( h ) = K exp( − t /τ ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making v = 0, and we get i ( p ) = V Th / R Th as a particular solution (which happens to be simply a constant). M. B. Patil, IIT Bombay

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), V Th = constant . * KVL: V Th = R Th i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L / R Th → i ( h ) = K exp( − t /τ ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making v = 0, and we get i ( p ) = V Th / R Th as a particular solution (which happens to be simply a constant). * i = i ( h ) + i ( p ) = K exp( − t /τ ) + V Th / R Th . M. B. Patil, IIT Bombay

RL circuits with DC sources R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), V Th = constant . * KVL: V Th = R Th i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L / R Th → i ( h ) = K exp( − t /τ ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making v = 0, and we get i ( p ) = V Th / R Th as a particular solution (which happens to be simply a constant). * i = i ( h ) + i ( p ) = K exp( − t /τ ) + V Th / R Th . * In general, i ( t ) = A exp( − t /τ ) + B , where A and B can be obtained from known conditions on i . M. B. Patil, IIT Bombay

RL circuits with DC sources (continued) R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), we have i ( t ) = A exp( − t /τ ) + B , τ = L / R Th . M. B. Patil, IIT Bombay

RL circuits with DC sources (continued) R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), we have i ( t ) = A exp( − t /τ ) + B , τ = L / R Th . � − 1 � * v ( t ) = L di ≡ A ′ exp( − t /τ ) . dt = L × A exp( − t /τ ) τ M. B. Patil, IIT Bombay

RL circuits with DC sources (continued) R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), we have i ( t ) = A exp( − t /τ ) + B , τ = L / R Th . � − 1 � * v ( t ) = L di ≡ A ′ exp( − t /τ ) . dt = L × A exp( − t /τ ) τ * As t → ∞ , v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish. M. B. Patil, IIT Bombay

RL circuits with DC sources (continued) R Th A A Circuit i i (resistors, ≡ voltage sources, V Th v L v L current sources, CCVS, CCCS, VCVS, VCCS) B B * If all sources are DC (constant), we have i ( t ) = A exp( − t /τ ) + B , τ = L / R Th . � − 1 � * v ( t ) = L di ≡ A ′ exp( − t /τ ) . dt = L × A exp( − t /τ ) τ * As t → ∞ , v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish. * Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can be expressed as x ( t ) = K 1 exp( − t /τ ) + K 2 , where K 1 and K 2 can be obtained from suitable conditions on x ( t ). M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V * V s changes from 0 V (at t = 0 − ), to 5 V (at t = 0 + ). As a result of this change, V c will rise. How fast can V c change? M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V * V s changes from 0 V (at t = 0 − ), to 5 V (at t = 0 + ). As a result of this change, V c will rise. How fast can V c change? * For example, what would happen if V c changes by 1 V in 1 µ s at a constant rate of 1 V / 1 µ s = 10 6 V / s? M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V * V s changes from 0 V (at t = 0 − ), to 5 V (at t = 0 + ). As a result of this change, V c will rise. How fast can V c change? * For example, what would happen if V c changes by 1 V in 1 µ s at a constant rate of 1 V / 1 µ s = 10 6 V / s? * i = C dV c = 1 µ F × 10 6 V s = 1 A . dt M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V * V s changes from 0 V (at t = 0 − ), to 5 V (at t = 0 + ). As a result of this change, V c will rise. How fast can V c change? * For example, what would happen if V c changes by 1 V in 1 µ s at a constant rate of 1 V / 1 µ s = 10 6 V / s? * i = C dV c = 1 µ F × 10 6 V s = 1 A . dt * With i = 1 A , the voltage drop across R would be 1000 V ! Not allowed by KVL. M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V * V s changes from 0 V (at t = 0 − ), to 5 V (at t = 0 + ). As a result of this change, V c will rise. How fast can V c change? * For example, what would happen if V c changes by 1 V in 1 µ s at a constant rate of 1 V / 1 µ s = 10 6 V / s? * i = C dV c = 1 µ F × 10 6 V s = 1 A . dt * With i = 1 A , the voltage drop across R would be 1000 V ! Not allowed by KVL. * We conclude that V c (0 + ) = V c (0 − ) ⇒ A capacitor does not allow abrupt changes in V c if there is a finite resistance in the circuit. M. B. Patil, IIT Bombay

RC circuits: Can V c change “suddenly?” R = 1 k Vs i 5 V Vs Vc C = 1 µ F Vc(0) = 0 V t 0 V * V s changes from 0 V (at t = 0 − ), to 5 V (at t = 0 + ). As a result of this change, V c will rise. How fast can V c change? * For example, what would happen if V c changes by 1 V in 1 µ s at a constant rate of 1 V / 1 µ s = 10 6 V / s? * i = C dV c = 1 µ F × 10 6 V s = 1 A . dt * With i = 1 A , the voltage drop across R would be 1000 V ! Not allowed by KVL. * We conclude that V c (0 + ) = V c (0 − ) ⇒ A capacitor does not allow abrupt changes in V c if there is a finite resistance in the circuit. * Similarly, an inductor does not allow abrupt changes in i L . M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : (1) v (0 − ) = Vs (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = 0 V Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : (1) v (0 − ) = Vs (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = 0 V Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t → ∞ : V 0 = B . i.e., B = V 0 , A = − V 0

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : (1) v (0 − ) = Vs (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = 0 V Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t → ∞ : V 0 = B . i.e., B = V 0 , A = − V 0 v ( t ) = V 0 [ 1 − exp ( − t /τ )]

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v v V s C V s C t t 0 V 0 V (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : (1) v (0 − ) = Vs (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = 0 V Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t → ∞ : V 0 = B . i.e., B = V 0 , A = − V 0 v ( t ) = V 0 [ 1 − exp ( − t /τ )]

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v v V s C V s C t t 0 V 0 V (A) (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : (1) v (0 − ) = Vs (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = 0 V Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t → ∞ : V 0 = B . i.e., B = V 0 , A = − V 0 v ( t ) = V 0 [ 1 − exp ( − t /τ )]

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v v V s C V s C t t 0 V 0 V (A) (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : Conditions on v(t) : (1) (1) v (0 − ) = Vs (0 − ) = 0 V v (0 − ) = Vs (0 − ) = V 0 v (0 + ) ≃ v (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = V 0 Note that we need the condition at 0 + (and not at 0 − ) Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . because Eq. (A) applies only for t > 0 . (2) (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = 0 V Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t → ∞ : V 0 = B . i.e., B = V 0 , A = − V 0 v ( t ) = V 0 [ 1 − exp ( − t /τ )]

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v v V s C V s C t t 0 V 0 V (A) (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : Conditions on v(t) : (1) (1) v (0 − ) = Vs (0 − ) = 0 V v (0 − ) = Vs (0 − ) = V 0 v (0 + ) ≃ v (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = V 0 Note that we need the condition at 0 + (and not at 0 − ) Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . because Eq. (A) applies only for t > 0 . (2) (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = 0 V Imposing (1) and (2) on Eq. (A), we get Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t = 0 + : V 0 = A + B , t → ∞ : V 0 = B . t → ∞ : 0 = B . i.e., B = V 0 , A = − V 0 i.e., A = V 0 , B = 0 v ( t ) = V 0 [ 1 − exp ( − t /τ )]

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v v V s C V s C t t 0 V 0 V (A) (A) Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Let v ( t ) = A exp ( − t /τ ) + B , t > 0 Conditions on v(t) : Conditions on v(t) : (1) (1) v (0 − ) = Vs (0 − ) = 0 V v (0 − ) = Vs (0 − ) = V 0 v (0 + ) ≃ v (0 − ) = 0 V v (0 + ) ≃ v (0 − ) = V 0 Note that we need the condition at 0 + (and not at 0 − ) Note that we need the condition at 0 + (and not at 0 − ) because Eq. (A) applies only for t > 0 . because Eq. (A) applies only for t > 0 . (2) (2) As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = V 0 As t → ∞ , i → 0 → v ( ∞ ) = Vs ( ∞ ) = 0 V Imposing (1) and (2) on Eq. (A), we get Imposing (1) and (2) on Eq. (A), we get t = 0 + : 0 = A + B , t = 0 + : V 0 = A + B , t → ∞ : V 0 = B . t → ∞ : 0 = B . i.e., B = V 0 , A = − V 0 i.e., A = V 0 , B = 0 v ( t ) = V 0 [ 1 − exp ( − t /τ )] v ( t ) = V 0 exp ( − t /τ ) M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V Compute i(t), t > 0 .

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V Compute i(t), t > 0 . i ( t ) = C d (A) dt V 0 [ 1 − exp ( − t /τ )] = CV 0 exp ( − t /τ ) = V 0 R exp ( − t /τ ) τ

RC circuits: charging and discharging transients R V s i V 0 v V s C t 0 V Compute i(t), t > 0 . i ( t ) = C d (A) dt V 0 [ 1 − exp ( − t /τ )] = CV 0 exp ( − t /τ ) = V 0 R exp ( − t /τ ) τ Let i ( t ) = A ′ exp ( − t /τ ) + B ′ , (B) t > 0 . t = 0 + : v = 0 , Vs = V 0 ⇒ i (0 + ) = V 0 / R . t → ∞ : i ( t ) = 0 . Using these conditions, we obtain A ′ = V 0 R , B ′ = 0 ⇒ i ( t ) = V 0 R exp ( − t /τ )

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v V s C V s v C t 0 V t 0 V Compute i(t), t > 0 . Compute i(t), t > 0 . i ( t ) = C d (A) dt V 0 [ 1 − exp ( − t /τ )] = CV 0 exp ( − t /τ ) = V 0 R exp ( − t /τ ) τ Let i ( t ) = A ′ exp ( − t /τ ) + B ′ , (B) t > 0 . t = 0 + : v = 0 , Vs = V 0 ⇒ i (0 + ) = V 0 / R . t → ∞ : i ( t ) = 0 . Using these conditions, we obtain A ′ = V 0 R , B ′ = 0 ⇒ i ( t ) = V 0 R exp ( − t /τ )

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v V s C V s v C t 0 V t 0 V Compute i(t), t > 0 . Compute i(t), t > 0 . i ( t ) = C d i ( t ) = C d (A) dt V 0 [ 1 − exp ( − t /τ )] (A) dt V 0 [ exp ( − t /τ )] = CV 0 exp ( − t /τ ) = V 0 = − CV 0 exp ( − t /τ ) = − V 0 R exp ( − t /τ ) R exp ( − t /τ ) τ τ Let i ( t ) = A ′ exp ( − t /τ ) + B ′ , (B) t > 0 . t = 0 + : v = 0 , Vs = V 0 ⇒ i (0 + ) = V 0 / R . t → ∞ : i ( t ) = 0 . Using these conditions, we obtain A ′ = V 0 R , B ′ = 0 ⇒ i ( t ) = V 0 R exp ( − t /τ )

RC circuits: charging and discharging transients R R V s V s i i V 0 V 0 v V s C V s v C t 0 V t 0 V Compute i(t), t > 0 . Compute i(t), t > 0 . i ( t ) = C d i ( t ) = C d (A) dt V 0 [ 1 − exp ( − t /τ )] (A) dt V 0 [ exp ( − t /τ )] = CV 0 exp ( − t /τ ) = V 0 = − CV 0 exp ( − t /τ ) = − V 0 R exp ( − t /τ ) R exp ( − t /τ ) τ τ Let i ( t ) = A ′ exp ( − t /τ ) + B ′ , Let i ( t ) = A ′ exp ( − t /τ ) + B ′ , (B) t > 0 . (B) t > 0 . t = 0 + : v = 0 , Vs = V 0 ⇒ i (0 + ) = V 0 / R . t = 0 + : v = V 0 , Vs = 0 ⇒ i (0 + ) = − V 0 / R . t → ∞ : i ( t ) = 0 . t → ∞ : i ( t ) = 0 . Using these conditions, we obtain Using these conditions, we obtain A ′ = V 0 R , B ′ = 0 ⇒ i ( t ) = V 0 A ′ = − V 0 R , B ′ = 0 ⇒ i ( t ) = − V 0 R exp ( − t /τ ) R exp ( − t /τ ) M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients R = 1 k Vs i 5 V Vs v C = 1 µ F t 0 V v ( t ) = V 0 [ 1 − exp ( − t /τ )] i ( t ) = V 0 R exp ( − t /τ )

RC circuits: charging and discharging transients R = 1 k Vs i 5 V Vs v C = 1 µ F t 0 V v ( t ) = V 0 [ 1 − exp ( − t /τ )] i ( t ) = V 0 R exp ( − t /τ ) 5 Vs v v (Volts) 0

RC circuits: charging and discharging transients R = 1 k Vs i 5 V Vs v C = 1 µ F t 0 V v ( t ) = V 0 [ 1 − exp ( − t /τ )] i ( t ) = V 0 R exp ( − t /τ ) 5 Vs v v (Volts) 0 5 i (mA) 0 −2 0 2 4 6 8 time (msec)

RC circuits: charging and discharging transients R = 1 k R = 1 k Vs Vs i i 5 V 5 V Vs v C = 1 µ F Vs v C 1 µ F t t 0 V 0 V v ( t ) = V 0 [ 1 − exp ( − t /τ )] v ( t ) = V 0 exp ( − t /τ ) i ( t ) = V 0 i ( t ) = − V 0 R exp ( − t /τ ) R exp ( − t /τ ) 5 Vs v v (Volts) 0 5 i (mA) 0 −2 0 2 4 6 8 time (msec)

RC circuits: charging and discharging transients R = 1 k R = 1 k Vs Vs i i 5 V 5 V Vs v C = 1 µ F Vs v C 1 µ F t t 0 V 0 V v ( t ) = V 0 [ 1 − exp ( − t /τ )] v ( t ) = V 0 exp ( − t /τ ) i ( t ) = V 0 i ( t ) = − V 0 R exp ( − t /τ ) R exp ( − t /τ ) 5 5 Vs Vs v v (Volts) v (Volts) v 0 0 5 i (mA) 0 −2 0 2 4 6 8 time (msec)

RC circuits: charging and discharging transients R = 1 k R = 1 k Vs Vs i i 5 V 5 V Vs v C = 1 µ F Vs v C 1 µ F t t 0 V 0 V v ( t ) = V 0 [ 1 − exp ( − t /τ )] v ( t ) = V 0 exp ( − t /τ ) i ( t ) = V 0 i ( t ) = − V 0 R exp ( − t /τ ) R exp ( − t /τ ) 5 5 Vs Vs v v (Volts) v (Volts) v 0 0 5 0 i (mA) i (mA) 0 −5 −2 0 2 4 6 8 −2 0 2 4 6 8 time (msec) time (msec) M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients R R Vs Vs i i 5 V 5 V Vs v C = 1 µ F Vs v C = 1 µ F 0 V t 0 V t v ( t ) = V 0 [ 1 − exp ( − t /τ )] v ( t ) = V 0 exp ( − t /τ ) R = 100 Ω 5 5 R = 1 k Ω v (Volts) v (Volts) R = 1 k Ω 0 0 R = 100 Ω −1 0 1 2 3 4 5 6 −1 0 1 2 3 4 5 6 time (msec) time (msec) M. B. Patil, IIT Bombay

Analysis of RC / RL circuits with a piece-wise constant source * Identify intervals in which the source voltages/currents are constant. For example, V s (1) t < t 1 (2) t 1 < t < t 2 (3) t > t 2 t 0 t 1 t 2 M. B. Patil, IIT Bombay

Analysis of RC / RL circuits with a piece-wise constant source * Identify intervals in which the source voltages/currents are constant. For example, V s (1) t < t 1 (2) t 1 < t < t 2 (3) t > t 2 t 0 t 1 t 2 * For any current or voltage x ( t ), write general expressions such as, x ( t ) = A 1 exp( − t /τ ) + B 1 , t < t 1 , x ( t ) = A 2 exp( − t /τ ) + B 2 , t 1 < t < t 2 , x ( t ) = A 3 exp( − t /τ ) + B 3 , t > t 2 . M. B. Patil, IIT Bombay

Analysis of RC / RL circuits with a piece-wise constant source * Identify intervals in which the source voltages/currents are constant. For example, V s (1) t < t 1 (2) t 1 < t < t 2 (3) t > t 2 t 0 t 1 t 2 * For any current or voltage x ( t ), write general expressions such as, x ( t ) = A 1 exp( − t /τ ) + B 1 , t < t 1 , x ( t ) = A 2 exp( − t /τ ) + B 2 , t 1 < t < t 2 , x ( t ) = A 3 exp( − t /τ ) + B 3 , t > t 2 . * Work out suitable conditions on x ( t ) at specific time points using M. B. Patil, IIT Bombay

Analysis of RC / RL circuits with a piece-wise constant source * Identify intervals in which the source voltages/currents are constant. For example, V s (1) t < t 1 (2) t 1 < t < t 2 (3) t > t 2 t 0 t 1 t 2 * For any current or voltage x ( t ), write general expressions such as, x ( t ) = A 1 exp( − t /τ ) + B 1 , t < t 1 , x ( t ) = A 2 exp( − t /τ ) + B 2 , t 1 < t < t 2 , x ( t ) = A 3 exp( − t /τ ) + B 3 , t > t 2 . * Work out suitable conditions on x ( t ) at specific time points using (a) If the source voltage/current has not changed for a “long” time (long compared to τ ), all derivatives are zero. ⇒ i C = C dV c = 0 , and V L = L di L dt = 0 . dt M. B. Patil, IIT Bombay

Analysis of RC / RL circuits with a piece-wise constant source * Identify intervals in which the source voltages/currents are constant. For example, V s (1) t < t 1 (2) t 1 < t < t 2 (3) t > t 2 t 0 t 1 t 2 * For any current or voltage x ( t ), write general expressions such as, x ( t ) = A 1 exp( − t /τ ) + B 1 , t < t 1 , x ( t ) = A 2 exp( − t /τ ) + B 2 , t 1 < t < t 2 , x ( t ) = A 3 exp( − t /τ ) + B 3 , t > t 2 . * Work out suitable conditions on x ( t ) at specific time points using (a) If the source voltage/current has not changed for a “long” time (long compared to τ ), all derivatives are zero. ⇒ i C = C dV c = 0 , and V L = L di L dt = 0 . dt (b) When a source voltage (or current) changes, say, at t = t 0 , V c ( t ) or i L ( t ) cannot change abruptly, i.e., V c ( t + 0 ) , and i L ( t + 0 ) = V c ( t − 0 ) = i L ( t − 0 ) . M. B. Patil, IIT Bombay

Analysis of RC / RL circuits with a piece-wise constant source * Identify intervals in which the source voltages/currents are constant. For example, V s (1) t < t 1 (2) t 1 < t < t 2 (3) t > t 2 t 0 t 1 t 2 * For any current or voltage x ( t ), write general expressions such as, x ( t ) = A 1 exp( − t /τ ) + B 1 , t < t 1 , x ( t ) = A 2 exp( − t /τ ) + B 2 , t 1 < t < t 2 , x ( t ) = A 3 exp( − t /τ ) + B 3 , t > t 2 . * Work out suitable conditions on x ( t ) at specific time points using (a) If the source voltage/current has not changed for a “long” time (long compared to τ ), all derivatives are zero. ⇒ i C = C dV c = 0 , and V L = L di L dt = 0 . dt (b) When a source voltage (or current) changes, say, at t = t 0 , V c ( t ) or i L ( t ) cannot change abruptly, i.e., V c ( t + 0 ) , and i L ( t + 0 ) = V c ( t − 0 ) = i L ( t − 0 ) . * Compute A 1 , B 1 , · · · using the conditions on x ( t ). M. B. Patil, IIT Bombay

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 Find i(t).

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 Find i(t). There are three intervals of constant V s : (1) t < t 0 (2) t 0 < t < t 1 (3) t > t 1

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 Find i(t). There are three intervals of constant V s : (1) t < t 0 (2) t 0 < t < t 1 (3) t > t 1 R Th seen by L is the same in all intervals: R Th = R 1 � R 2 = 8 Ω R 1 τ = L / R Th R 2 V s = 0.8 H / 8 Ω = 0.1 s

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 Find i(t). At t = t − 0 , v = 0 V, V s = 0 V . There are three intervals of constant V s : ⇒ i ( t − 0 ) = 0 A ⇒ i ( t + 0 ) = 0 A . (1) t < t 0 (2) t 0 < t < t 1 (3) t > t 1 R Th seen by L is the same in all intervals: R Th = R 1 � R 2 = 8 Ω R 1 τ = L / R Th R 2 V s = 0.8 H / 8 Ω = 0.1 s

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 Find i(t). At t = t − 0 , v = 0 V, V s = 0 V . There are three intervals of constant V s : ⇒ i ( t − 0 ) = 0 A ⇒ i ( t + 0 ) = 0 A . (1) t < t 0 If V s did not change at t = t 1 , (2) t 0 < t < t 1 we would have (3) t > t 1 V s R Th seen by L is the same in all intervals: 10 V R Th = R 1 � R 2 = 8 Ω R 1 t t 0 t 1 τ = L / R Th R 2 V s = 0.8 H / 8 Ω v ( ∞ ) = 0 V , i ( ∞ ) = 10 V / 10 Ω = 1 A . = 0.1 s

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 Find i(t). At t = t − 0 , v = 0 V, V s = 0 V . There are three intervals of constant V s : ⇒ i ( t − 0 ) = 0 A ⇒ i ( t + 0 ) = 0 A . (1) t < t 0 If V s did not change at t = t 1 , (2) t 0 < t < t 1 we would have (3) t > t 1 V s R Th seen by L is the same in all intervals: 10 V R Th = R 1 � R 2 = 8 Ω R 1 t t 0 t 1 τ = L / R Th R 2 V s = 0.8 H / 8 Ω v ( ∞ ) = 0 V , i ( ∞ ) = 10 V / 10 Ω = 1 A . = 0.1 s Using i ( t + 0 ) and i ( ∞ ) , we can obtain i ( t ) , t > 0 (See next slide). M. B. Patil, IIT Bombay

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 1 i (Amp) 0 0 0.2 0.4 0.6 0.8 time (sec)

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 1 i (Amp) 0 0 0.2 0.4 0.6 0.8 time (sec) In reality, V s changes at t = t 1 , and we need to work out the solution for t > t 1 separately.

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 For t 0 < t < t 1 , i ( t ) = 1 − exp ( − t /τ ) Amp . 1 Consider t > t 1 . i (Amp) 0 0 0.2 0.4 0.6 0.8 time (sec) In reality, V s changes at t = t 1 , and we need to work out the solution for t > t 1 separately.

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 For t 0 < t < t 1 , i ( t ) = 1 − exp ( − t /τ ) Amp . 1 Consider t > t 1 . 1 ) = 1 − e − 1 = 0 . 632 A (Note: t 1 /τ = 1). i ( t + 1 ) = i ( t − i (Amp) i ( ∞ ) = 0 A . 0 0 0.2 0.4 0.6 0.8 time (sec) In reality, V s changes at t = t 1 , and we need to work out the solution for t > t 1 separately.

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 For t 0 < t < t 1 , i ( t ) = 1 − exp ( − t /τ ) Amp . 1 Consider t > t 1 . 1 ) = 1 − e − 1 = 0 . 632 A (Note: t 1 /τ = 1). i ( t + 1 ) = i ( t − i (Amp) i ( ∞ ) = 0 A . Let i ( t ) = A exp ( − t /τ ) + B. 0 0 0.2 0.4 0.6 0.8 time (sec) In reality, V s changes at t = t 1 , and we need to work out the solution for t > t 1 separately.

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 For t 0 < t < t 1 , i ( t ) = 1 − exp ( − t /τ ) Amp . 1 Consider t > t 1 . 1 ) = 1 − e − 1 = 0 . 632 A (Note: t 1 /τ = 1). i ( t + 1 ) = i ( t − i (Amp) i ( ∞ ) = 0 A . Let i ( t ) = A exp ( − t /τ ) + B. 0 It is convenient to rewrite i ( t ) as i ( t ) = A ′ exp [ − ( t − t 1 ) /τ ] + B. 0 0.2 0.4 0.6 0.8 time (sec) In reality, V s changes at t = t 1 , and we need to work out the solution for t > t 1 separately.

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H R 2 V s v t 0 = 0 t t 1 = 0.1 s t 0 t 1 For t 0 < t < t 1 , i ( t ) = 1 − exp ( − t /τ ) Amp . 1 Consider t > t 1 . 1 ) = 1 − e − 1 = 0 . 632 A (Note: t 1 /τ = 1). i ( t + 1 ) = i ( t − i (Amp) i ( ∞ ) = 0 A . Let i ( t ) = A exp ( − t /τ ) + B. 0 It is convenient to rewrite i ( t ) as i ( t ) = A ′ exp [ − ( t − t 1 ) /τ ] + B. 0 0.2 0.4 0.6 0.8 time (sec) Using i ( t + 1 ) and i ( ∞ ) , we get In reality, V s changes at t = t 1 , i ( t ) = 0 . 632 exp [ − ( t − t 1 ) /τ ] A. and we need to work out the solution for t > t 1 separately. M. B. Patil, IIT Bombay

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 i ( t ) = 0 . 632 exp [ − ( t − t 1 ) /τ ] A. 1 i (Amp) 0 0 0.2 0.4 0.6 0.8 time (sec)

RL circuit: example R 1 R 1 = 10 Ω V s R 2 = 40 Ω i 10 V L = 0 . 8 H V s R 2 v t 0 = 0 t t 1 = 0.1 s t 0 t 1 i ( t ) = 0 . 632 exp [ − ( t − t 1 ) /τ ] A. Combining the solutions for t 0 < t < t 1 and t > t 1 , we get 1 1 i (Amp) i (Amp) 0 0 0.2 0.4 0.6 0.8 0 time (sec) 0 0.2 0.4 0.6 0.8 time (sec) (SEQUEL file: ee101_rl1.sqproj) M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k t=0 i i c 1 k 5 k v c R 2 R 1 5 µ F 6 V

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA .

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V ⇒ i ( 0 + ) = 5 V / ( 5 k + 5 k ) = 0.5 mA.

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V ⇒ i ( 0 + ) = 5 V / ( 5 k + 5 k ) = 0.5 mA. Let i(t) = A exp(-t /τ ) + B for t > 0, with τ = 10 k × 5 µ F = 50 ms.

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V ⇒ i ( 0 + ) = 5 V / ( 5 k + 5 k ) = 0.5 mA. Let i(t) = A exp(-t /τ ) + B for t > 0, with τ = 10 k × 5 µ F = 50 ms. Using i ( 0 + ) and i ( ∞ ) = 0 A, we get i(t) = 0.5 exp(-t /τ ) mA.

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V ⇒ i ( 0 + ) = 5 V / ( 5 k + 5 k ) = 0.5 mA. Let i(t) = A exp(-t /τ ) + B for t > 0, with τ = 10 k × 5 µ F = 50 ms. Using i ( 0 + ) and i ( ∞ ) = 0 A, we get i(t) = 0.5 exp(-t /τ ) mA. 1 i (mA) 0 0 0.5 time (sec)

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V ⇒ i ( 0 + ) = 5 V / ( 5 k + 5 k ) = 0.5 mA. Let i(t) = A exp(-t /τ ) + B for t > 0, with τ = 10 k × 5 µ F = 50 ms. Using i ( 0 + ) and i ( ∞ ) = 0 A, we get i(t) = 0.5 exp(-t /τ ) mA. 1 5 0 i c (mA) i (mA) v c (V) 0 0 0.5 −0.5 0 0 0.5 0 0.5 time (sec) time (sec) time (sec)

RC circuit: The switch has been closed for a long time and opens at t = 0. R 3 = 5 k 5 k 5 k t=0 i i c i i c i i c 1 k 1 k 5 k 5 k 5 k v c v c v c R 2 AND R 1 5 µ F 5 µ F 5 µ F 6 V 6 V t < 0 t > 0 t = 0 − : capacitor is an open circuit ⇒ i ( 0 − ) = 6 V / ( 5 k + 1 k ) = 1 mA . v c (0 − ) = i(0 − ) R 1 = 5 V ⇒ v c (0 + ) = 5 V ⇒ i ( 0 + ) = 5 V / ( 5 k + 5 k ) = 0.5 mA. Let i(t) = A exp(-t /τ ) + B for t > 0, with τ = 10 k × 5 µ F = 50 ms. Using i ( 0 + ) and i ( ∞ ) = 0 A, we get (SEQUEL file: ee101_rc2.sqproj) i(t) = 0.5 exp(-t /τ ) mA. 1 5 0 i c (mA) i (mA) v c (V) 0 0 0.5 −0.5 0 0 0.5 0 0.5 time (sec) time (sec) time (sec) M. B. Patil, IIT Bombay

RC circuit: example i C R 10 5 k Vc V S C V S , V C (Volts) 1 µ F 0 2 1 i C (mA) 0 − 1 0 10 20 30 40 50 t (msec) M. B. Patil, IIT Bombay

RC circuit: example i C R 10 5 k Vc V S C V S , V C (Volts) 1 µ F Find expressions for V C ( t ) and i C ( t ) in steady state (in terms of R , C , V 0 , T 1 , 0 T 2 ). 2 V 0 1 i C (mA) 0 0 T 1 T 2 − 1 0 10 20 30 40 50 t (msec) M. B. Patil, IIT Bombay