Problem 161 - A Complete Resource Book in Physics for JEE Main

Question

Two sources of equal EMF are connected to an external resistance \(R\). The internal resistances of the two sources are \(R_{1}\) and \(R_{2}\left(R_{2}>R_{1}\right)\). If the potential difference across the source having internal resistance \(R_{2}\) is zero, then (A) \(R=\frac{R_{2} \times\left(R_{1}+R_{2}\right)}{\left(R_{2}-R_{1}\right)}\) (B) \(R=R_{2}-R_{1}\) (C) \(R=\frac{R_{1} R_{2}}{\left(R_{1}+R_{2}\right)}\) (D) \(R=\frac{R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\)

Step-by-Step Solution

Answer

The short answer to the question is: \(R=\frac{R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\).

Step 1: Understand the circuit

In this exercise, we have two equal EMF sources (call them \(\varepsilon\)) connected in a parallel manner with an external resistance \(R\). Each source has its internal resistance, source 1 has \(R_{1}\) and source 2 has \(R_{2}\). The potential difference across source 2 with internal resistance \(R_{2}\) is zero.

Step 2: Apply Kirchhoff's loop rule

Kirchhoff's loop rule states that the sum of the voltage changes (losses or gains) in a closed loop in a circuit is zero. Let's apply this rule for both loops, starting from point A (assuming clockwise direction): Loop 1: \(-\varepsilon + I_{1}R_{1} + I_{1}R = 0\) Loop 2: \(-\varepsilon + I_{2}R_{2} + I_{2}R = 0\)

Step 3: Calculate current and potential difference

The potential difference across the source with internal resistance \(R_{2}\) is zero. We can write this as: \(V_{R2} = I_{2}R_{2}\) Since the potential difference across source 1 is equal to the potential difference across source 2, we can write the following equation: \(I_{1}R_{1} + I_{1}R = I_{2}R_{2}\)

Step 4: Express \(I_1\) and \(I_2\) in terms of \(R\), \(R_1\), and \(R_2\)

From the loop rules: \(I_{1}=\frac{\varepsilon}{R_{1}+R}\) and \(I_{2}=\frac{\varepsilon}{R_{2}+R}\) Replace \(I_{1}\) and \(I_{2}\) in our equation: \[\frac{\varepsilon}{R_{1}+R}R_{1} + \frac{\varepsilon}{R_{1}+R}R = \frac{\varepsilon}{R_{2}+R}R_{2}\]

Step 5: Solve for external resistance \(R\)

First, we can cancel out the \(\varepsilon\) in the equation, then rearrange the equation and solve for \(R\): \[R_{1}(R_{1}+R) + R(R_{1}+R) = R_{2}(R_{2}+R)\] Expanding the equation and collecting the \(R\) terms: \[R^{2} + 2R_{1}R + R_{1}^{2} = R^{2} + (R_{2}^{2} - R_{1}^{2})\] Rearranging and solving for \(R\): \[R = \frac{R_{1}R_{2}}{R_{2} - R_{1}}\] So, the correct answer is: (D) \(R=\frac{R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\)