The stored charge in the capacitor in steady state of the following circuit is ______ $\mu$C.
Numerical value type. Enter your answer.
Answer: 200
In steady state no current flows through the capacitor branch, so the $10\ \Omega$ in series with it carries no current and the capacitor voltage equals the voltage across the $4\ \Omega$ vertical resistor.
Reduce the ladder from the right:
- $4\ \Omega$ (top) $+ 4\ \Omega$ (vertical) $+ 2\ \Omega$ (bottom) $= 10\ \Omega$, in parallel with the $10\ \Omega$ vertical $\Rightarrow 5\ \Omega$.
- $5\ \Omega$ (top) $+ 5\ \Omega + 2\ \Omega$ (bottom) $= 12\ \Omega$. This is directly across the $12$ V battery (in parallel with the $12\ \Omega$), so it carries $1$ A.
Voltage across the $10\ \Omega$ vertical $= 12 - 1 \times 7 = 5$ V. Current in the right branch $= \dfrac{5}{10} = 0.5$ A, so the $4\ \Omega$ vertical has $2$ V.
$Q = CV = 100\ \mu\text{F} \times 2\ \text{V} = 200\ \mu$C.
Solution by Sreeraj P, M.Sc Physics