JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 19)
A coil of self inductance 10 mH and resistance 0.1 $$\Omega $$ is connected through a switch to a battery of internal
resistance 0.9 $$\Omega $$. After the switch is closed, the time taken for the current to attain 80% of the saturation
value is: [take ln 5 = 1.6]
0.324 s
0.002 s
0.103 s
0.016 s
Explanation
L = 10 × 10–3 H, r1 = 0.1 $$\Omega $$
$$i = \varepsilon \left\{ {1 - {e^{ - 1/2}}} \right\}$$
$${i_{saturation}}{\rm{ }} = {\rm{ }}\varepsilon $$
$$80\% {\rm{ }}{i_{saturation}}{\rm{ }} = {\rm{ }}0.8{\rm{ }}\varepsilon $$
0.8 = 1 - e-t/2 ; e-t/2 = 0.2
et/L = 5
t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3
$$i = \varepsilon \left\{ {1 - {e^{ - 1/2}}} \right\}$$
$${i_{saturation}}{\rm{ }} = {\rm{ }}\varepsilon $$
$$80\% {\rm{ }}{i_{saturation}}{\rm{ }} = {\rm{ }}0.8{\rm{ }}\varepsilon $$
0.8 = 1 - e-t/2 ; e-t/2 = 0.2
et/L = 5
t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3
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