Question

Find the charge on the capacitor in the steady state. \vspace{1cm} \includegraphics[width=0.7\textwidth]{a.png}

Answer 1

Given the circuit with an 8 µF capacitor, resistances of 10Ω and 15Ω, and voltage sources of 5V and 8V, we need to find the charge on the capacitor in the steady state. In the steady state, the capacitor behaves as an open circuit because the current cannot flow through it anymore (after a long time). So, we can simplify the circuit by ignoring the capacitor and finding the equivalent voltage across it. - First, calculate the total resistance \( R_{\text{total}} \). The resistances of 10Ω and 15Ω are in series, so: \[ R_{\text{total}} = 10Ω + 15Ω = 25Ω. \] - Now, calculate the total voltage \( V_{\text{total}} \). The two voltage sources (5V and 8V) are in series, so: \[ V_{\text{total}} = 5V + 8V = 13V. \] - Using Ohm's law, \( V = IR \), we can find the current \( I \) in the circuit: \[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{13V}{25Ω} = 0.52A. \] - In the steady state, the charge \( Q \) on the capacitor is given by: \[ Q = C \times V_{\text{capacitor}}, \] where \( C = 8 \, \mu\text{F} \) and \( V_{\text{capacitor}} \) is the voltage across the capacitor. In the steady state, the voltage across the capacitor is the same as the voltage across the 10Ω resistor, which is: \[ V_{\text{capacitor}} = I \times 10Ω = 0.52A \times 10Ω = 5.2V. \] Therefore, the charge on the capacitor is: \[ Q = 8 \, \mu\text{F} \times 5.2V = 8 \times 10^{-6} \, \text{F} \times 5.2V = 41.6 \, \mu\text{C}. \] Thus, the charge on the capacitor is \( 41.6 \, \mu\text{C} \).