In the given circuit, if the current through 4 $\Omega$ resistor is zero, then the value of the resistance $R$ is:
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When the current through a resistor in a circuit is zero, the potential difference across it is zero. This often indicates a balance condition, similar to a Wheatstone bridge, where the ratio of voltages across parallel branches can help find unknown resistances.
\textbf{Step 1: Analyze the Circuit}
The given circuit consists of:
A $ 9 \, \text{V} $ battery connected to a $ 1 \, \Omega $ resistor.
A $ 6 \, \text{V} $ battery connected in series with a $ 4 \, \Omega $ resistor.
A $ 3 \, \text{V} $ battery connected in series with a variable resistor $ R $.
We are told that the current through the $ 4 \, \Omega $ resistor is zero. This implies that no current flows through the branch containing the $ 4 \, \Omega $ resistor, meaning the potential difference across it must be zero. Therefore, the voltage at both ends of the $ 4 \, \Omega $ resistor must be equal.
Step 2: Apply Kirchhoff's Voltage Law (KVL)
To ensure the current through the $ 4 \, \Omega $ resistor is zero, the voltages at the junctions must balance such that no potential difference exists across the $ 4 \, \Omega $ resistor. Let's analyze the loop involving the $ 4 \, \Omega $ resistor and the $ R $ resistor.
Loop Analysis
1. Consider the loop involving the $ 6 \, \text{V} $ battery, $ 4 \, \Omega $ resistor, and $ R $ resistor.
2. For the current through the $ 4 \, \Omega $ resistor to be zero, the voltage drop across it must be zero. This means the voltage at the top and bottom of the $ 4 \, \Omega $ resistor must be the same.
Step 3: Use Voltage Division Principle
Since the current through the $ 4 \, \Omega $ resistor is zero, the voltage across it is zero. This implies that the voltage at the junction where the $ 4 \, \Omega $ resistor connects to the $ R $ resistor must be the same as the voltage at the other end of the $ 4 \, \Omega $ resistor.
Voltage at Junctions
The voltage at the top of the $ 4 \, \Omega $ resistor is determined by the $ 6 \, \text{V} $ battery.
The voltage at the bottom of the $ 4 \, \Omega $ resistor is determined by the $ 3 \, \text{V} $ battery and the resistance $ R $.
For the current through the $ 4 \, \Omega $ resistor to be zero, the voltage drop across it must be zero. This means the voltage at the junction must satisfy:
$$
V_{\text{top}} = V_{\text{bottom}}
$$
Calculate the Equivalent Resistance
To find the value of $ R $, we use the condition that the voltage across the $ 4 \, \Omega $ resistor is zero. This implies that the resistances must be balanced such that the voltage division ensures no current flows through the $ 4 \, \Omega $ resistor.
Using the principle of voltage division:
$$
\frac{R}{R + 4} = \frac{3}{6}
$$
Simplify:
$$
\frac{R}{R + 4} = \frac{1}{2}
$$
Cross-multiply:
$$
2R = R + 4
$$
Solve for $ R $:
$$
R = 4 - R
$$
$$
R = 2 \, \Omega
$$
Step 4: Verify the Solution
If $ R = 2 \, \Omega $, the voltage division ensures that the voltage across the $ 4 \, \Omega $ resistor is zero, confirming that no current flows through it.
Step 5: Analyze the Options
Option (1): $ 1 \, \Omega $
Incorrect — does not satisfy the condition for zero current through the $ 4 \, \Omega $ resistor.
Option (2): $ 2 \, \Omega $
Correct — matches the calculated value.
Option (3): $ 3 \, \Omega $
Incorrect — does not satisfy the condition for zero current through the $ 4 \, \Omega $ resistor.
Option (4): $ 4 \, \Omega $
Incorrect — does not satisfy the condition for zero current through the $ 4 \, \Omega $ resistor.
