CUET · MATHS · PYQ PAPER 2024
A cell of emf 1.1 V and internal resistance \(0.5 \Omega\) is connected to a \(0.5 \Omega\) wire. Another cell of same emf is added in series, but current remains unchanged. The second cell's internal resistance is :
- A \(1 \Omega\)
- B \(2.5 \Omega\)
- C \(1.5 \Omega\)
- D \(2 \Omega\)
Answer & Solution
Correct Answer
(A) \(1 \Omega\)
Step-by-step Solution
Detailed explanation
\( \frac{E_1}{R+r_1} = \frac{E_1+E_2}{R+r_1+r_2} \) \( \frac{1.1}{0.5+0.5} = \frac{1.1+1.1}{0.5+0.5+r_2} \) \( \frac{1.1}{1} = \frac{2.2}{1+r_2} \) \( 1+r_2 = 2 \) \( r_2 = 1 \Omega \)
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