Current Electricity

JEE Main level, numerical answer, medium

202 of 380

Question 202 of 380

The stored charge in the capacitor in steady state of the following circuit is ________ μ\muC.

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Answer: 200

In steady state, the capacitor acts as an open circuit, meaning no current flows through the branch containing the 100 μF100\ \mu\text{F} capacitor. Consequently, no current flows through the 10 Ω10\ \Omega resistor in series with it on the top right.

The voltage across the capacitor will be equal to the voltage across the 4 Ω4\ \Omega vertical resistor.

Let's simplify the circuit to find this voltage. The 12 Ω12\ \Omega resistor on the left is in parallel with the 12 V12\ \text{V} battery, so the voltage across the rest of the circuit remains 12 V12\ \text{V}.

For the rightmost part of the active circuit, the 4 Ω4\ \Omega top resistor, the 4 Ω4\ \Omega vertical resistor, and the 2 Ω2\ \Omega bottom resistor are in series. Their equivalent resistance is:
Rright=4 Ω+4 Ω+2 Ω=10 ΩR_{\text{right}} = 4\ \Omega + 4\ \Omega + 2\ \Omega = 10\ \Omega

This 10 Ω10\ \Omega equivalent resistance is in parallel with the 10 Ω10\ \Omega vertical resistor in the middle. The equivalent resistance of this parallel combination is:
Rmid=10×1010+10=5 ΩR_{\text{mid}} = \dfrac{10 \times 10}{10 + 10} = 5\ \Omega

Now, the total resistance of the main circuit connected to the 12 V12\ \text{V} source is the sum of the 5 Ω5\ \Omega top-left resistor, the 5 Ω5\ \Omega equivalent middle resistance, and the 2 Ω2\ \Omega bottom-left resistor:
Rtotal=5 Ω+5 Ω+2 Ω=12 ΩR_{\text{total}} = 5\ \Omega + 5\ \Omega + 2\ \Omega = 12\ \Omega

The total current flowing from the 12 V12\ \text{V} source into this main branch is:
I=12 V12 Ω=1 AI = \dfrac{12\ \text{V}}{12\ \Omega} = 1\ \text{A}

The voltage across the parallel combination (the middle section) is:
Vmid=I×Rmid=1 A×5 Ω=5 VV_{\text{mid}} = I \times R_{\text{mid}} = 1\ \text{A} \times 5\ \Omega = 5\ \text{V}

This 5 V5\ \text{V} is applied across the right branch which has a total resistance of 10 Ω10\ \Omega. The current through this branch is:
Iright=5 V10 Ω=0.5 AI_{\text{right}} = \dfrac{5\ \text{V}}{10\ \Omega} = 0.5\ \text{A}

The voltage across the 4 Ω4\ \Omega vertical resistor is:
V4Ω=Iright×4 Ω=0.5 A×4 Ω=2 VV_{4\Omega} = I_{\text{right}} \times 4\ \Omega = 0.5\ \text{A} \times 4\ \Omega = 2\ \text{V}

Since the capacitor is in parallel with this 4 Ω4\ \Omega resistor (with no voltage drop across the 10 Ω10\ \Omega resistor in its branch), the voltage across the capacitor is also 2 V2\ \text{V}.

The stored charge in the capacitor is:
Q=C×V=100 μF×2 V=200 μCQ = C \times V = 100\ \mu\text{F} \times 2\ \text{V} = 200\ \mu\text{C}

Answer: 200200

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