Question

Figure below shows a network of resistors, cells, and a capacitor at steady state.
 
What is the current through the resistance 4 Ω?

• A. 1.0 A
• B. 0.2 A
• C. Zero
• D. 0.5 A

Hint:

Remember that at steady state, the capacitor behaves like an open circuit and does not influence the current. Simplify the circuit accordingly for easier analysis.

Answer 1

The Correct Option is B

In this circuit, the capacitor has reached a steady state. At steady state, a capacitor behaves like an open circuit because there is no current flowing through it. Therefore, the capacitor is irrelevant to the current calculations once steady state is reached. We are asked to find the current through the 4Ω resistor.
1. Simplify the circuit: The 2Ω resistor and the 6V battery are in series. The equivalent resistance of the top portion of the circuit (2Ω) is in parallel with the 1Ω and 2Ω resistors connected in series (to form a 3Ω resistance).
2. The total resistance of the circuit can be found by calculating the equivalent resistance of the parallel combination and then adding the remaining resistors. First, let's combine the 1Ω and 2Ω resistors: \[ R_{\text{eq}} = 1Ω + 2Ω = 3Ω \] This 3Ω is now in parallel with the 2Ω resistor at the top of the circuit. The equivalent resistance of the parallel combination is: \[ R_{\text{parallel}} = \frac{2Ω \times 3Ω}{2Ω + 3Ω} = \frac{6Ω}{5Ω} = 1.2Ω \] Now, this 1.2Ω is in series with the 4Ω resistor and the 3V battery, so the total resistance in the circuit is: \[ R_{\text{total}} = 1.2Ω + 4Ω = 5.2Ω \]
3. Using Ohm's Law, the total current \( I_{\text{total}} \) is: \[ I_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{6V + 3V}{5.2Ω} = \frac{9V}{5.2Ω} = 1.73 \, \text{A} \]
4. The current through the 4Ω resistor can be calculated by using the current division rule. The current divides between the 1.2Ω and 4Ω resistors. The current through the 4Ω resistor is: \[ I_{4Ω} = I_{\text{total}} \times \frac{R_{\text{parallel}}}{R_{\text{parallel}} + 4Ω} = 1.73 \, \text{A} \times \frac{1.2Ω}{1.2Ω + 4Ω} = 1.73 \, \text{A} \times \frac{1.2Ω}{5.2Ω} \] \[ I_{4Ω} = 1.73 \times 0.231 = 0.2 \, \text{A} \]
Thus, the current through the 4Ω resistor is 0.2 A.