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A copper wire of radius \(0.1 \mathrm{~mm}\) and resistance \(2 \mathrm{k} \Omega\) is connected across a power supply of \(40 \mathrm{~V}\). The number of electrons transferred per second between the supply and the wire at one end is
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Verified Answer
The correct answer is:
\(1.25 \times 10^{17}\)
Radius of copper wire, \(r=0.1 \mathrm{~mm}\)
\(=1 \times 10^{-4} \mathrm{~m}\)
Resistance, \(R=2 \mathrm{k} \Omega=2 \times 10^3 \Omega\)
Power supply, \(V=40 \mathrm{~V}\)
Current flowing through the wire,
\(I=\frac{V}{R}=\frac{40}{2 \times 10^3}=2 \times 10^{-2} \mathrm{~A}\)
\(\therefore\) Charge flowing per second
\(q=I t=2 \times 10^{-2} \times 1=2 \times 10^{-2} \mathrm{C}\)
\(\therefore\) Number of electrons transferred,
\(\begin{aligned}
n & =\frac{q}{e}=\frac{2 \times 10^{-2}}{1.6 \times 10^{-19}} \\
& =1.25 \times 10^{17} \text { electrons }
\end{aligned}\)
\(=1 \times 10^{-4} \mathrm{~m}\)
Resistance, \(R=2 \mathrm{k} \Omega=2 \times 10^3 \Omega\)
Power supply, \(V=40 \mathrm{~V}\)
Current flowing through the wire,
\(I=\frac{V}{R}=\frac{40}{2 \times 10^3}=2 \times 10^{-2} \mathrm{~A}\)
\(\therefore\) Charge flowing per second
\(q=I t=2 \times 10^{-2} \times 1=2 \times 10^{-2} \mathrm{C}\)
\(\therefore\) Number of electrons transferred,
\(\begin{aligned}
n & =\frac{q}{e}=\frac{2 \times 10^{-2}}{1.6 \times 10^{-19}} \\
& =1.25 \times 10^{17} \text { electrons }
\end{aligned}\)
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