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Circuit Theory

RC transients-2

In the circui...

Question

In the circuit shown, the initial voltages across the capacitors $C_{1}$ and $C_{2}$ are 1 V and 3 V respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached, is .

1.5

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Solution

The correct option is A 1.5

$I n i t i a l e n e r g y = \frac{1}{2} (C_{1} V_{1}^{2} + C_{2} V_{2}^{2})$

$= \frac{1}{2} (3 \times 1^{2} + 1 \times 3^{2}) = 6 J$

Final energy stored in capacitor
$= \frac{1}{2} (C_{1} + C_{2}) V^{2}$

$C_{1} V_{1} + C_{2} V_{2} = (C_{1} + C_{2}) V$

$1 \times 3 + 3 \times 1 = (1 + 3) V$

$V = 1.5 V$

$F i n a l e n e r g y = \frac{1}{2} (1 + 3) \times (1.5)^{2} = 4.5 J$
$E n e r g y d i s s i p a t e d = 6 - 4.5 = 1.5 J$

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In the circuit shown, the initial voltages across the capacitors C1 and C2 are 1 V and 3 V respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached, is . 1.5

The correct option is A 1.5 Initial energy=12(C1V21+C2V22) =12(3×12+1×32)=6 J Final energy stored in capacitor =12(C1+C2)V2 C1V1+C2V2=(C1+C2)V 1×3+3×1=(1+3)V V=1.5 V Final energy=12(1+3)×(1.5)2=4.5 J Energy dissipated=6−4.5=1.5J

In the circuit shown, the initial voltages across the capacitors $C_{1}$ and $C_{2}$ are 1 V and 3 V respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached, is .

1.5