Question:
Under what conditions current passing through the resistance \(R\) can be increased by short circuiting the battery of emf \(E_2\)? The internal resistances of the two batteries are \(r_1\) and \(r_2\) respectively.
Under what conditions current passing through the resistance \(R\) can be increased by short circuiting the battery of emf \(E_2\)? The internal resistances of the two batteries are \(r_1\) and \(r_2\) respectively.
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Always derive the condition carefully. The inequality direction depends on what is being compared.
Updated On: Apr 20, 2026
- \(E_2 r_1>E_1 (R + r_2)\)
- \(E_1 r_2<E_2 (r_1 + R)\)
- \(E_2 r_2<E_1 (R + r_2)\)
- \(E_1 r_1>E_2 (r_1 + R)\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Compare current in \(R\) in two cases: with both batteries in circuit, and when \(E_2\) is short-circuited.
Step 2: Detailed Explanation:
Current with both batteries: \(I_1 = \frac{E_1 + E_2}{R + r_1 + r_2}\). Current with \(E_2\) shorted: \(I_2 = \frac{E_1}{R + r_1}\). Condition: \(I_2>I_1 \Rightarrow \frac{E_1}{R+r_1}>\frac{E_1+E_2}{R+r_1+r_2}\). Cross multiplication: \(E_1(R+r_1+r_2)>(E_1+E_2)(R+r_1)\). Simplifying: \(E_1 r_2>E_2(R+r_1)\).
Step 3: Final Answer:
\[ \boxed{E_1 r_2<E_2 (r_1 + R)} \]
Compare current in \(R\) in two cases: with both batteries in circuit, and when \(E_2\) is short-circuited.
Step 2: Detailed Explanation:
Current with both batteries: \(I_1 = \frac{E_1 + E_2}{R + r_1 + r_2}\). Current with \(E_2\) shorted: \(I_2 = \frac{E_1}{R + r_1}\). Condition: \(I_2>I_1 \Rightarrow \frac{E_1}{R+r_1}>\frac{E_1+E_2}{R+r_1+r_2}\). Cross multiplication: \(E_1(R+r_1+r_2)>(E_1+E_2)(R+r_1)\). Simplifying: \(E_1 r_2>E_2(R+r_1)\).
Step 3: Final Answer:
\[ \boxed{E_1 r_2<E_2 (r_1 + R)} \]
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