Question

A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 77.9% of its final value? (b) What is the current through the resistor at a time t = 1.00τL?

Answer 1

Answer:

(A) $9.14\times 10^{-9}sec$

(B) $6.20\times 10^{-3}A$

Explanation:

We have given inductance $L=5.41\mu H=5.41\times 10^{-6}H$

Resistance $R=0.949kohm=0.949\times 10^3ohm$

Time constant of RL circuit is equal to $\tau =\frac{L}{R}$

$\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec$

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

$i=i_0(1-e^{\frac{-t}{\tau }})$

According to question

$0.799i_0=i_0(1-e^{\frac{-t}{\tau }})$

$e^{\frac{-t}{\tau }}=0.201$

${\frac{-t}{\tau }}=ln0.201$

${\frac{-t}{5.7\times 10^{-9} }}=-1.6044$

$t=9.14\times 10^{-9}sec$

(b) Current at $t=\tau sec$

$i=i_0(1-e^{\frac{-t}{\tau }})$

$i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})$

$i=6.20\times 10^{-3}A$