Answer :
Explanation:
To find the potential difference across the resistance \(2R\), let's analyze the circuit.
We have two resistors in parallel: \(4R\) and \(2R\), and these are in series with \(R\) connected to a voltage source \(E\).
First, let's find the equivalent resistance of the parallel combination of \(4R\) and \(2R\). The formula for the equivalent resistance of resistors in parallel is:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{4R} + \frac{1}{2R} \]
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{4R} + \frac{2}{4R} \]
\[ \frac{1}{R_{\text{eq}}} = \frac{3}{4R} \]
\[ R_{\text{eq}} = \frac{4R}{3} \]
Now, the total resistance in the circuit is the sum of \(R_{\text{eq}}\) and \(R\):
\[ R_{\text{total}} = R_{\text{eq}} + R \]
\[ R_{\text{total}} = \frac{4R}{3} + R \]
\[ R_{\text{total}} = \frac{4R + 3R}{3} \]
\[ R_{\text{total}} = \frac{7R}{3} \]
Now, we can use Ohm's Law to find the potential difference across \(2R\):
\[ V = IR \]
But first, we need to find the current flowing through the circuit. We can use Ohm's Law again:
\[ I = \frac{V}{R_{\text{total}}} \]
\[ I = \frac{E}{\frac{7R}{3}} \]
\[ I = \frac{3E}{7R} \]
Now, we can find the potential difference across \(2R\):
\[ V_{2R} = IR \]
\[ V_{2R} = \left(\frac{3E}{7R}\right) \cdot 2R \]
\[ V_{2R} = \frac{6E}{7} \]
So, the potential difference across the resistance \(2R\) is \(\frac{6E}{7}\), which is option (A).