Question

The stored charge in the capacitor in steady state of the following circuit is ________ µC.

Answer 1

Correct Answer: 4

Step 1: Behaviour of capacitor in steady state In steady state for a DC circuit, the capacitor behaves as an open circuit. Therefore, no current flows through the capacitor branch. So we first find the potential difference across the capacitor terminals using resistor reduction. Step 2: Simplify the rightmost section The rightmost resistors \(10\Omega\), \(4\Omega\), and \(2\Omega\) are in series: \[ R_{\text{right}}=10+4+2=16\Omega \] This is in parallel with the \(10\Omega\) vertical resistor: \[ R_{eq2}=\frac{16\times 10}{16+10} \] \[ R_{eq2}=\frac{160}{26} \] \[ R_{eq2}=\frac{80}{13}\Omega \] Step 3: Move one step left This equivalent resistance is in series with \(4\Omega\) and \(2\Omega\): \[ R_{\text{branch}}=4+\frac{80}{13}+2 \] \[ R_{\text{branch}}=6+\frac{80}{13} \] \[ R_{\text{branch}}=\frac{158}{13}\Omega \] Now this branch is in parallel with the \(12\Omega\) resistor: \[ R_{eq1}= \frac{12\cdot \frac{158}{13}} {12+\frac{158}{13}} \] \[ R_{eq1}= \frac{1896/13}{314/13} \] \[ R_{eq1}=\frac{948}{157}\Omega \] Step 4: Total circuit resistance Including the \(5\Omega\) series resistor: \[ R_{\text{total}}=5+\frac{948}{157} \] \[ R_{\text{total}}=\frac{1733}{157}\Omega \] Hence total current from the \(12V\) battery is \[ I=\frac{12}{R_{\text{total}}} \] \[ I=12\cdot \frac{157}{1733}\text{ A} \] Step 5: Voltage across capacitor Voltage across first parallel section: \[ V_1=I\cdot R_{eq1} \] \[ V_1= \left(12\cdot \frac{157}{1733}\right) \cdot \frac{948}{157} \] \[ V_1= 12\cdot \frac{948}{1733} \] Voltage across second section: \[ V_2= V_1\cdot \frac{R_{eq2}}{R_{\text{branch}}} \] \[ V_2= V_1\cdot \frac{80/13}{158/13} \] \[ V_2= V_1\cdot \frac{40}{79} \] Since capacitor is across the \(4\Omega\) resistor of the \(16\Omega\) branch: \[ V_C= V_2\cdot \frac{4}{16} \] \[ V_C=\frac{V_2}{4} \] Substituting: \[ V_C= \frac{1}{4}\cdot \left(12\cdot \frac{948}{1733}\right)\cdot \frac{40}{79} \] \[ V_C\approx 0.8309\text{ V} \] Step 6: Charge stored Using \[ Q=CV \] Given \[ C=100\,\mu F \] \[ Q=100\times 0.8309 \] \[ Q=83.09\,\mu C \] \[ Q\approx 83\,\mu C \] Final Answer: \[ \boxed{83\,\mu C} \]