Find the voltage and current passing through the resistor $R_2$ shown in the following circuit
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Always treat capacitors in DC circuits as open paths when analyzing steady-state current. This simplifies the network into a standard resistive circuit.
Concept:
In a DC circuit containing a capacitor, once the capacitor is fully charged, it acts as an open circuit (steady-state condition). This means no steady current flows through the branch containing the capacitor.
The total resistance in a series circuit is $R_{total} = R_1 + R_2$, and the current is $I = \frac{V}{R_{total}}$.
Step 1: {Analyze the circuit in steady-state.}
In steady-state, the 10 pF capacitor acts as an open circuit. Current from the 3 V battery flows only through the series combination of resistors $R_1$ and $R_2$.
Step 2: {Calculate the total current in the circuit.}
Given $R_1 = 1 \text{ k}\Omega$ and $R_2 = 2 \text{ k}\Omega$ connected in series:
$$R_{total} = R_1 + R_2 = 1 \text{ k}\Omega + 2 \text{ k}\Omega = 3 \text{ k}\Omega$$
The total current $I$ is:
$$I = \frac{V}{R_{total}} = \frac{3 \text{ V}}{3 \text{ k}\Omega} = 1 \text{ mA}$$
Step 3: {Find the voltage across resistor $R_2$.}
Since $R_2$ is in series, the current passing through it is $1 \text{ mA}$. The voltage $V_2$ across it is:
$$V_2 = I \times R_2 = 1 \text{ mA} \times 2 \text{ k}\Omega = 2 \text{ V} \dots \text{}$$
If $R_2$ is the horizontal resistor shown as $2 \text{ k}\Omega$ and the branch splits, the voltage depends on position. Based on the options and standard loop analysis, $I = 1 \text{ mA}$ and $V = 1 \text{ V}$ corresponds to $R_1$ or a parallel segment.
