COMEDK · Physics · 19. Current Electricity
A cell of emf 5 V and negligible internal resistance is connected across a conducting wire of cross-sectional area \(2 \mathrm{~mm}^2\). If the electron density in the wire is \(5 \times 10^{28}\) and the drift velocity of the electrons is \(0.125 \mathrm{mms}^{-1}\) then the resistance offered by the wire is:
- A \(2.5 \Omega\)
- B \(0.4 \Omega\)
- C \(5 \Omega\)
- D \(4 \Omega\)
Answer & Solution
Correct Answer
(A) \(2.5 \Omega\)
Step-by-step Solution
Detailed explanation
The current \(I\) in a conductor is given by the relation \(I = nAev_d\), where \(n\) is the electron density, \(A\) is the cross-sectional area, \(e\) is the elementary charge, and \(v_d\) is the drift velocity. Given values are: \(n = 5 \times 10^{28} \text{ m}^{-3}\)…
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