KCET · Physics · Alternating Current
The current following through an inductance coil of self-inductance \(6 \mathrm{mH}\) at different time instants is as shown. The emf induced between \(t=20 \mathrm{~s}\) and \(t=40 \mathrm{~s}\) is nearly
- A \(2 \times 10^{-2} \mathrm{~V}\)
- B \(3 \times 10^{-2} \mathrm{~V}\)
- C \(4 \times 10^{-3} \mathrm{~V}\)
- D \(30 \times 10^{-4} \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(30 \times 10^{-4} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
As we know,
Induced emf, \(|e|=L \frac{d I}{d t}\)
Here, \(L=6 \mathrm{mH}=6 \times 10^{-3} \mathrm{H}\)
From the given graph,
\(\begin{aligned}
|e| &=L \frac{\left(I_{2}-I_{1}\right)}{\left(t_{2}-t_{1}\right)} \\
&=\frac{L}{20}=\frac{6 \times 10^{-3}}{20}=3 \times 10^{-4} \mathrm{~V}
\end{aligned}\)
Induced emf, \(|e|=L \frac{d I}{d t}\)
Here, \(L=6 \mathrm{mH}=6 \times 10^{-3} \mathrm{H}\)
From the given graph,
\(\begin{aligned}
|e| &=L \frac{\left(I_{2}-I_{1}\right)}{\left(t_{2}-t_{1}\right)} \\
&=\frac{L}{20}=\frac{6 \times 10^{-3}}{20}=3 \times 10^{-4} \mathrm{~V}
\end{aligned}\)
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