The stored charge in the capacitor in steady state of the following circuit is 2.5em0.4pt. Numerical answer type. Work it out, then open the answer.

Answer: 200. Steady state: capacitor branch carries no current. Right part: 4+4+2 = 10 Ω in parallel with the middle 10 Ω gives 5 Ω. Total with 5 Ω and 2 Ω: 12 Ω across 12 V, so I = 1 A and 5 V across the middle 10 Ω. Current in the right path = 0.5 A, so the right 4 Ω has 0.5×4 = 2 V. Capacitor is across that 4 Ω: Q = 100×2 = 200.

The stored charge in the capacitor in steady state of the following… | JEE Main 2026 (8 Apr, Shift 2)

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Capacitors in DC Circuits & RC

The stored charge in the capacitor in steady state of the following circuit is $\rule{2.5em}{0.4pt}\ \mu\text{C}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 200

Steady state: capacitor branch carries no current. Right part: $4+4+2 = 10\ \Omega$ in parallel with the middle $10\ \Omega$ gives $5\ \Omega$ .
Total with $5\ \Omega$ and $2\ \Omega$ : $12\ \Omega$ across 12 V, so $I = 1\text{ A}$ and 5 V across the middle $10\ \Omega$ .
Current in the right path $= 0.5\text{ A}$ , so the right $4\ \Omega$ has $0.5\times4 = 2\text{ V}$ .
Capacitor is across that $4\ \Omega$ : $Q = 100\times2 = 200\ \mu\text{C}$ .

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