MHT CET · Physics · Current Electricity
A \(10 \mathrm{~m}\) long wire of resistance \(20 \Omega\) is connected in series with a battery of e.m.f. \(3 \mathrm{~V}\) (negligible internal resistance) and a resistance of \(10 \Omega\). The potential gradient along the wire is
- A \(3 \mathrm{~V} / \mathrm{m}\)
- B \(0.1 \mathrm{~V} / \mathrm{m}\)
- C \(0.2 \mathrm{~V} / \mathrm{m}\)
- D \(0.3 \mathrm{~V} / \mathrm{m}\)
Answer & Solution
Correct Answer
(C) \(0.2 \mathrm{~V} / \mathrm{m}\)
Step-by-step Solution
Detailed explanation
Two resistances are in series
\(\mathrm{R}_{\mathrm{net}}=20 \Omega+10 \Omega=30 \Omega\)
Current \(\mathrm{I}=\frac{3}{30}=0.1 \mathrm{~A}\)
Potential through the wire is \(\mathrm{V}=0.1 \times 20=2 \mathrm{~V}\)
\(\therefore \quad\) Potential gradient \(=\frac{\mathrm{V}}{10}=\frac{2}{10}=0.2 \frac{\mathrm{V}}{\mathrm{m}}\)
\(\mathrm{R}_{\mathrm{net}}=20 \Omega+10 \Omega=30 \Omega\)
Current \(\mathrm{I}=\frac{3}{30}=0.1 \mathrm{~A}\)
Potential through the wire is \(\mathrm{V}=0.1 \times 20=2 \mathrm{~V}\)
\(\therefore \quad\) Potential gradient \(=\frac{\mathrm{V}}{10}=\frac{2}{10}=0.2 \frac{\mathrm{V}}{\mathrm{m}}\)
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