Question

Find the currents in the resistors \(2 \, \Omega\) and \(10 \, \Omega\) in the network shown in figure, also find charge on the capacitor :
\includegraphics[width=0.5\linewidth]{2.png}

Hint:

In any DC circuit problem with capacitors, always analyze the circuit at \(t=0\) (capacitor is a short circuit) and \(t \to \infty\) (capacitor is an open circuit). The question usually implies the steady-state condition unless it asks about the transient phase.

Answer 1

Step 1: Understanding the Concept (Steady State in DC Circuit):
This is a DC circuit containing a capacitor. When a DC voltage is applied, the capacitor charges up. After a long time, it becomes fully charged and its acts as an open circuit, meaning no current can flow through the branch containing the capacitor. This is called the steady state.

Step 2: Circuit Analysis at Steady State:
\begin{enumerate} \item Current in the 10 \(\Omega\) resistor: At steady state, the capacitor \(C = 10 \, \mu\text{F}\) is fully charged and blocks the flow of DC current. Therefore, the current in the middle branch, which contains the \(10 \, \Omega\) resistor, becomes zero. \[ I_{10\Omega} = 0 \text{ A} \] \item Current in the 2 \(\Omega\) resistor: Since no current flows through the middle branch, the circuit simplifies to a single series loop consisting of the \(2\) V battery, the \(2 \, \Omega\) resistor, and the \(8 \, \Omega\) resistor.
The total resistance in this loop is: \[ R_{total} = 2 \, \Omega + 8 \, \Omega = 10 \, \Omega \] Using Ohm's law, the current flowing through this loop is: \[ I_{2\Omega} = \frac{V}{R_{total}} = \frac{2 \text{ V}}{10 \, \Omega} = 0.2 \text{ A} \] This is the current flowing through both the \(2 \, \Omega\) and \(8 \, \Omega\) resistors.
\item Charge on the Capacitor: To find the charge \(Q\) on the capacitor, we first need to find the potential difference (voltage) \(V_C\) across it. The capacitor and the 10 \(\Omega\) resistor are connected in parallel with the \(8 \, \Omega\) resistor (based on the common interpretation of this circuit's drawing). Therefore, the voltage across the capacitor branch is the same as the voltage across the \(8 \, \Omega\) resistor. \[ V_C = V_{8\Omega} \] The voltage across the \(8 \, \Omega\) resistor can be calculated using Ohm's law, with the current we found in the previous step: \[ V_{8\Omega} = I_{8\Omega} \times R_{8\Omega} = (0.2 \text{ A}) \times (8 \, \Omega) = 1.6 \text{ V} \] So, the voltage across the capacitor is \(V_C = 1.6\) V. Note that because the current through the 10 \(\Omega\) resistor is zero, there is no voltage drop across it, so the full 1.6 V appears across the capacitor plates.
The charge \(Q\) on the capacitor is given by \(Q = C V_C\): \[ Q = (10 \, \mu\text{F}) \times (1.6 \text{ V}) = 16 \, \mu\text{C} \] \end{enumerate}

Step 3: Final Answer:
\begin{itemize} \item Current in the \(2 \, \Omega\) resistor = 0.2 A.
\item Current in the \(10 \, \Omega\) resistor = 0 A.
\item Charge on the capacitor = 16 \(\mu\)C.
\end{itemize}