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Question The RC circuit shown below has a variable resistance
R ( t ) R(t) R ( t ) given by the following expression:
R ( t ) = R 0 ( 1 − t T ) for 0 ≤ t < T R(t) = R_0 \left( 1 - \frac{t}{T} \right) \text{ for } 0 \leq t < T R ( t ) = R 0 ( 1 − T t ) for 0 ≤ t < T where
R 0 = 1 Ω R_0 = 1\text{ }\Omega R 0 = 1 Ω , and
C = 1 F C = 1\text{ F} C = 1 F . We are also given that
T = 3 R 0 C T = 3 R_0 C T = 3 R 0 C and the source voltage is
V s = 1 V V_s = 1\text{ V} V s = 1 V . If the current at time
t = 0 t = 0 t = 0 is
1 A 1\text{ A} 1 A , then the current
I ( t ) I(t) I ( t ) , in amperes, at time
t = T / 2 t = T/2 t = T /2 is
________ (rounded off to 2 decimal places).
Correct answer 0.23 to 0.27
Solution The KVL equation for the circuit for
t ≥ 0 t \geq 0 t ≥ 0 is:
V s = I ( t ) R ( t ) + V c ( t ) V_s = I(t)R(t) + V_c(t) V s = I ( t ) R ( t ) + V c ( t ) Given
I ( t ) = C d V c ( t ) d t I(t) = C \frac{dV_c(t)}{dt} I ( t ) = C d t d V c ( t ) , we have:
V s = R ( t ) C d V c ( t ) d t + V c ( t ) V_s = R(t) C \frac{dV_c(t)}{dt} + V_c(t) V s = R ( t ) C d t d V c ( t ) + V c ( t ) Substituting
R ( t ) = R 0 ( 1 − t / T ) R(t) = R_0 (1 - t/T) R ( t ) = R 0 ( 1 − t / T ) and
V s = 1 V_s = 1 V s = 1 :
1 = R 0 ( 1 − t / T ) C d V c d t + V c 1 = R_0 (1 - t/T) C \frac{dV_c}{dt} + V_c 1 = R 0 ( 1 − t / T ) C d t d V c + V c d V c d t + V c R 0 C ( 1 − t / T ) = 1 R 0 C ( 1 − t / T ) \frac{dV_c}{dt} + \frac{V_c}{R_0 C (1 - t/T)} = \frac{1}{R_0 C (1 - t/T)} d t d V c + R 0 C ( 1 − t / T ) V c = R 0 C ( 1 − t / T ) 1 This is a first-order linear ODE. The integrating factor is:
I F = e ∫ 1 R 0 C ( 1 − t / T ) d t = e − T R 0 C ln ( 1 − t / T ) = ( 1 − t / T ) − T / R 0 C IF = e^{\int \frac{1}{R_0 C (1 - t/T)} dt} = e^{-\frac{T}{R_0 C} \ln(1 - t/T)} = (1 - t/T)^{-T/R_0 C} I F = e ∫ R 0 C ( 1 − t / T ) 1 d t = e − R 0 C T l n ( 1 − t / T ) = ( 1 − t / T ) − T / R 0 C Given
T = 3 R 0 C T = 3 R_0 C T = 3 R 0 C , so
T / R 0 C = 3 T/R_0 C = 3 T / R 0 C = 3 . Thus,
I F = ( 1 − t / T ) − 3 IF = (1 - t/T)^{-3} I F = ( 1 − t / T ) − 3 .
Multiplying the ODE by the IF and integrating:
d d t [ V c ( t ) ( 1 − t / T ) − 3 ] = 1 R 0 C ( 1 − t / T ) − 4 \frac{d}{dt} [V_c(t) (1 - t/T)^{-3}] = \frac{1}{R_0 C} (1 - t/T)^{-4} d t d [ V c ( t ) ( 1 − t / T ) − 3 ] = R 0 C 1 ( 1 − t / T ) − 4 V c ( t ) ( 1 − t / T ) − 3 = 1 R 0 C [ ( 1 − t / T ) − 3 − 3 ( − 1 / T ) ] + K = T 3 R 0 C ( 1 − t / T ) − 3 + K V_c(t) (1 - t/T)^{-3} = \frac{1}{R_0 C} \left[ \frac{(1 - t/T)^{-3}}{-3 (-1/T)} \right] + K = \frac{T}{3 R_0 C} (1 - t/T)^{-3} + K V c ( t ) ( 1 − t / T ) − 3 = R 0 C 1 [ − 3 ( − 1/ T ) ( 1 − t / T ) − 3 ] + K = 3 R 0 C T ( 1 − t / T ) − 3 + K Since
T = 3 R 0 C T = 3 R_0 C T = 3 R 0 C , we get
V c ( t ) = 1 + K ( 1 − t / T ) 3 V_c(t) = 1 + K(1 - t/T)^3 V c ( t ) = 1 + K ( 1 − t / T ) 3 .
At
t = 0 t=0 t = 0 ,
I ( 0 ) = 1 A I(0) = 1\text{ A} I ( 0 ) = 1 A and
R ( 0 ) = R 0 = 1 Ω R(0) = R_0 = 1\text{ }\Omega R ( 0 ) = R 0 = 1 Ω . From
V s = I ( 0 ) R ( 0 ) + V c ( 0 ) V_s = I(0)R(0) + V_c(0) V s = I ( 0 ) R ( 0 ) + V c ( 0 ) , we find
1 = 1 ( 1 ) + V c ( 0 ) ⇒ V c ( 0 ) = 0 V 1 = 1(1) + V_c(0) \Rightarrow V_c(0) = 0\text{ V} 1 = 1 ( 1 ) + V c ( 0 ) ⇒ V c ( 0 ) = 0 V .
0 = 1 + K ( 1 − 0 ) 3 ⇒ K = − 1 0 = 1 + K(1 - 0)^3 \Rightarrow K = -1 0 = 1 + K ( 1 − 0 ) 3 ⇒ K = − 1 . So,
V c ( t ) = 1 − ( 1 − t / T ) 3 V_c(t) = 1 - (1 - t/T)^3 V c ( t ) = 1 − ( 1 − t / T ) 3 .
The current is
I ( t ) = C d V c d t = C [ − 3 ( 1 − t / T ) 2 ( − 1 / T ) ] = 3 C T ( 1 − t / T ) 2 I(t) = C \frac{dV_c}{dt} = C \left[ -3(1 - t/T)^2 (-1/T) \right] = \frac{3C}{T} (1 - t/T)^2 I ( t ) = C d t d V c = C [ − 3 ( 1 − t / T ) 2 ( − 1/ T ) ] = T 3 C ( 1 − t / T ) 2 .
At
t = T / 2 t = T/2 t = T /2 :
I ( T / 2 ) = 3 C T ( 1 − 1 / 2 ) 2 = 3 C 4 T = 3 C 4 ( 3 R 0 C ) = 1 4 R 0 = 1 4 ( 1 ) = 0.25 A I(T/2) = \frac{3C}{T} (1 - 1/2)^2 = \frac{3C}{4T} = \frac{3C}{4(3 R_0 C)} = \frac{1}{4 R_0} = \frac{1}{4(1)} = 0.25\text{ A} I ( T /2 ) = T 3 C ( 1 − 1/2 ) 2 = 4 T 3 C = 4 ( 3 R 0 C ) 3 C = 4 R 0 1 = 4 ( 1 ) 1 = 0.25 A .
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Correct answer 0.23 to 0.27
Solution The KVL equation for the circuit for
t ≥ 0 t \geq 0 t ≥ 0 is:
V s = I ( t ) R ( t ) + V c ( t ) V_s = I(t)R(t) + V_c(t) V s = I ( t ) R ( t ) + V c ( t ) Given
I ( t ) = C d V c ( t ) d t I(t) = C \frac{dV_c(t)}{dt} I ( t ) = C d t d V c ( t ) , we have:
V s = R ( t ) C d V c ( t ) d t + V c ( t ) V_s = R(t) C \frac{dV_c(t)}{dt} + V_c(t) V s = R ( t ) C d t d V c ( t ) + V c ( t ) Substituting
R ( t ) = R 0 ( 1 − t / T ) R(t) = R_0 (1 - t/T) R ( t ) = R 0 ( 1 − t / T ) and
V s = 1 V_s = 1 V s = 1 :
1 = R 0 ( 1 − t / T ) C d V c d t + V c 1 = R_0 (1 - t/T) C \frac{dV_c}{dt} + V_c 1 = R 0 ( 1 − t / T ) C d t d V c + V c d V c d t + V c R 0 C ( 1 − t / T ) = 1 R 0 C ( 1 − t / T ) \frac{dV_c}{dt} + \frac{V_c}{R_0 C (1 - t/T)} = \frac{1}{R_0 C (1 - t/T)} d t d V c + R 0 C ( 1 − t / T ) V c = R 0 C ( 1 − t / T ) 1 This is a first-order linear ODE. The integrating factor is:
I F = e ∫ 1 R 0 C ( 1 − t / T ) d t = e − T R 0 C ln ( 1 − t / T ) = ( 1 − t / T ) − T / R 0 C IF = e^{\int \frac{1}{R_0 C (1 - t/T)} dt} = e^{-\frac{T}{R_0 C} \ln(1 - t/T)} = (1 - t/T)^{-T/R_0 C} I F = e ∫ R 0 C ( 1 − t / T ) 1 d t = e − R 0 C T l n ( 1 − t / T ) = ( 1 − t / T ) − T / R 0 C Given
T = 3 R 0 C T = 3 R_0 C T = 3 R 0 C , so
T / R 0 C = 3 T/R_0 C = 3 T / R 0 C = 3 . Thus,
I F = ( 1 − t / T ) − 3 IF = (1 - t/T)^{-3} I F = ( 1 − t / T ) − 3 .
Multiplying the ODE by the IF and integrating:
d d t [ V c ( t ) ( 1 − t / T ) − 3 ] = 1 R 0 C ( 1 − t / T ) − 4 \frac{d}{dt} [V_c(t) (1 - t/T)^{-3}] = \frac{1}{R_0 C} (1 - t/T)^{-4} d t d [ V c ( t ) ( 1 − t / T ) − 3 ] = R 0 C 1 ( 1 − t / T ) − 4 V c ( t ) ( 1 − t / T ) − 3 = 1 R 0 C [ ( 1 − t / T ) − 3 − 3 ( − 1 / T ) ] + K = T 3 R 0 C ( 1 − t / T ) − 3 + K V_c(t) (1 - t/T)^{-3} = \frac{1}{R_0 C} \left[ \frac{(1 - t/T)^{-3}}{-3 (-1/T)} \right] + K = \frac{T}{3 R_0 C} (1 - t/T)^{-3} + K V c ( t ) ( 1 − t / T ) − 3 = R 0 C 1 [ − 3 ( − 1/ T ) ( 1 − t / T ) − 3 ] + K = 3 R 0 C T ( 1 − t / T ) − 3 + K Since
T = 3 R 0 C T = 3 R_0 C T = 3 R 0 C , we get
V c ( t ) = 1 + K ( 1 − t / T ) 3 V_c(t) = 1 + K(1 - t/T)^3 V c ( t ) = 1 + K ( 1 − t / T ) 3 .
At
t = 0 t=0 t = 0 ,
I ( 0 ) = 1 A I(0) = 1\text{ A} I ( 0 ) = 1 A and
R ( 0 ) = R 0 = 1 Ω R(0) = R_0 = 1\text{ }\Omega R ( 0 ) = R 0 = 1 Ω . From
V s = I ( 0 ) R ( 0 ) + V c ( 0 ) V_s = I(0)R(0) + V_c(0) V s = I ( 0 ) R ( 0 ) + V c ( 0 ) , we find
1 = 1 ( 1 ) + V c ( 0 ) ⇒ V c ( 0 ) = 0 V 1 = 1(1) + V_c(0) \Rightarrow V_c(0) = 0\text{ V} 1 = 1 ( 1 ) + V c ( 0 ) ⇒ V c ( 0 ) = 0 V .
0 = 1 + K ( 1 − 0 ) 3 ⇒ K = − 1 0 = 1 + K(1 - 0)^3 \Rightarrow K = -1 0 = 1 + K ( 1 − 0 ) 3 ⇒ K = − 1 . So,
V c ( t ) = 1 − ( 1 − t / T ) 3 V_c(t) = 1 - (1 - t/T)^3 V c ( t ) = 1 − ( 1 − t / T ) 3 .
The current is
I ( t ) = C d V c d t = C [ − 3 ( 1 − t / T ) 2 ( − 1 / T ) ] = 3 C T ( 1 − t / T ) 2 I(t) = C \frac{dV_c}{dt} = C \left[ -3(1 - t/T)^2 (-1/T) \right] = \frac{3C}{T} (1 - t/T)^2 I ( t ) = C d t d V c = C [ − 3 ( 1 − t / T ) 2 ( − 1/ T ) ] = T 3 C ( 1 − t / T ) 2 .
At
t = T / 2 t = T/2 t = T /2 :
I ( T / 2 ) = 3 C T ( 1 − 1 / 2 ) 2 = 3 C 4 T = 3 C 4 ( 3 R 0 C ) = 1 4 R 0 = 1 4 ( 1 ) = 0.25 A I(T/2) = \frac{3C}{T} (1 - 1/2)^2 = \frac{3C}{4T} = \frac{3C}{4(3 R_0 C)} = \frac{1}{4 R_0} = \frac{1}{4(1)} = 0.25\text{ A} I ( T /2 ) = T 3 C ( 1 − 1/2 ) 2 = 4 T 3 C = 4 ( 3 R 0 C ) 3 C = 4 R 0 1 = 4 ( 1 ) 1 = 0.25 A .
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