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Chapter 4: Q24E (page 1014)

In Fig. 30.11,R=15惟and the battery emf is 6.3V. With switch S2open, switch S1is closed. After several minutes, S1is opened and S2is closed. (a) At 2msafter S1is opened, the current has decayed to 0.28A. Calculate the inductance of the coil. (b) How long after S1is opened will the current reach 1%of its original value?

Short Answer

Expert verified

a)L=0.074H

b)t=0.0227s

Step by step solution

01

Calculate the inductance using the formula of current

The current in an RL circuit is measured by i=I0e-RLt, where I0=蔚R.

e-RLt=iR蔚e-RLt=0.28脳156.3e-RLt=0.6667

Taking natural logarithm on both the sides.

-RLt=ln0.6667L=-Rtln0.6667L=-15脳2脳10-3In0.6667L=0.074H

02

Calculate the time for  decay

For 1%decay, iI0=0.01.

From the equation of current,

e-RLt=0.01-RLt=ln0.01t=-ln0.01脳LRt=-ln0.01脳0.07415t=0.0227s

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