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Question: A 10 V battery connected to 5 \(\Omega\) resistance coil having inductance 10 H through a switch dri...

A 10 V battery connected to 5 Ω\Omega resistance coil having inductance 10 H through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is 2 ms. The average emf induced across the coil is

A

4 × 104 V

B

2 × 104 V

C

2 × 102 V

D

1 × 104 V

Answer

1 × 104 V

Explanation

Solution

Here, current I=VR=105=2 A\mathrm { I } = \frac { \mathrm { V } } { \mathrm { R } } = \frac { 10 } { 5 } = 2 \mathrm {~A}

Final current, when the switch is opened is zero

As θ=LdIdt\theta = - \mathrm { L } \frac { \mathrm { dI } } { \mathrm { dt } }

=10(1×103)=104 V= - 10 \left( - 1 \times 10 ^ { 3 } \right) = 10 ^ { 4 } \mathrm {~V}