CUET · PHYSICS · PYQ PAPER 2025
A battery of emf 4 V and internal resistance \(1 \, \Omega\) is connected across a wire of resistance \(9.0 \, \Omega\). The number of electrons passing through the cross-section of the wire in 4 s is:
- A \(10^{19}\)
- B \(10^{20}\)
- C \(10^{18}\)
- D \(2.5 \times 10^{18}\)
Answer & Solution
Correct Answer
(A) \(10^{19}\)
Step-by-step Solution
Detailed explanation
\(R_{eq} = 9.0 \\, \\Omega + 1 \\, \\Omega = 10 \\, \\Omega\) \(I = \frac{4 \\, V}{10 \\, \\Omega} = 0.4 \\, A\) \(n = \frac{I t}{e} = \frac{0.4 \\, A \\times 4 \\, s}{1.6 \\times 10^{-19} \\, C} = 1 \\times 10^{19}\)
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