JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A \(10\, \Omega, 20 \; mH\) coil carrying constant current is connected to a battery of \(20 \; V\) through a switch is opened current becomes zero in \(100 \; \mu s\). The average emf induced in the coil is \(\dots \; V\).
- A \(100\)
- B \(200\)
- C \(300\)
- D \(400\)
Answer & Solution
Correct Answer
(D) \(400\)
Step-by-step Solution
Detailed explanation
\(<\varepsilon>=\frac{\int \varepsilon dt }{\int dt }=\frac{\int( Ldi / dt ) dt }{\int dt }=\frac{ L \int di }{\int dt }\) \(<\varepsilon>=\frac{ L \Delta i }{\Delta i }\) \(i _{0}=\frac{ V }{ R }=\frac{20}{10}=2 \; A , \text { if } i =0 \; A\) \(T =100 \; \mu s , L =20 \; mH\)…
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