Question 217: AC fundamentals

Question

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]

Accepted Answer

L = 10 ÃÂ 10
Ã¢ÂÂ3
H, r
1
= 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
e
t/L
= 5
t = L ln 5 = 10 ÃÂ 10
Ã¢ÂÂ3
ÃÂ 1.6 = 16 ÃÂ 10
Ã¢ÂÂ3

Answer

0.324 s

Answer

0.002 s

Answer

0.103 s

Answer

0.016 s

AC fundamentals

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]

Choose the correct answer: