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Chapter 4: Q26E (page 1014)

In Fig.3.011, switchS1is closed while switchS2is kept open. The inductance isL=0.115H, and the resistance isR=120惟. (a) When the current has reached its final value, the energy stored in the inductor is0.26J. What is the emf蔚of the battery? (b) After the current has reached its final value,S1is opened androle="math" localid="1664253376215" S2is closed. How much time does it take for the energy stored in the inductor to decrease to0.13J, half of the original value?

Short Answer

Expert verified

a) 蔚=256V

b)role="math" localid="1664253462958" t=3.32脳10-4s

Step by step solution

01

Calculate the EMF using energy and current equations

The current equation is given by i=imax(1-e-tT)and the energy equation is given by U=12Li2.

Given that L=0.115H,R=120惟and U=0.26J.

U=12Li2i=2ULi=2脳0.260.115i=2.13A

Use resistance and current to find EMF.

蔚=iR蔚=2.13脳120蔚=256V

02

Calculate time for inductor energy to become half

The decreasing current is given by i=imaxe-tT

.The energy stored in the inductor will be given by

U=12Li2U=12Li2maxe-2RLtU=Umaxe-2RLt

Now, the energy becomes half the maximum value.

UUmax=12=e-2RLt

Take natural logarithm on both the sides.

localid="1664254817377" t=-L2Rln12t=0.115脳ln22脳120t=3.32脳10-4s

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