Question:
A dc source of emf \(E_1=100\) V and internal resistance \(r=0.5\ \Omega\), a storage battery of emf \(E_2=90\) V and an external resistance R are connected as shown in figure. For what value of R no current will pass through the battery? 
A dc source of emf \(E_1=100\) V and internal resistance \(r=0.5\ \Omega\), a storage battery of emf \(E_2=90\) V and an external resistance R are connected as shown in figure. For what value of R no current will pass through the battery?
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The three branches are in parallel. No current in the \(E_2\) branch means the voltage across R (and across the \(E_1\) branch terminals) equals \(E_2\).
Updated On: Oct 31, 2026
- \(5.5\ \Omega\)
- \(3.5\ \Omega\)
- \(4.5\ \Omega\)
- \(2.5\ \Omega\)
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The Correct Option is C
Solution and Explanation
With no current in the \(E_2\) branch, the circuit is just \(E_1\), \(r\) and R in series. Current \(I=\frac{E_1}{r+R}\).
The voltage across the parallel branches equals \(IR\), and this must equal \(E_2\) for zero current through the battery: \(\frac{100R}{0.5+R}=90\).
\(100R=45+90R\Rightarrow10R=45\Rightarrow R=4.5\ \Omega\).
Answer: (3) \(4.5\ \Omega\)
The voltage across the parallel branches equals \(IR\), and this must equal \(E_2\) for zero current through the battery: \(\frac{100R}{0.5+R}=90\).
\(100R=45+90R\Rightarrow10R=45\Rightarrow R=4.5\ \Omega\).
Answer: (3) \(4.5\ \Omega\)
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