In Fig. 30.11, suppose that ε=60V, R=240Ω, and L=0.16H. With switch S2open, switch S1is left closed until a constant current is established. Then S1is closed and S2opened, taking the battery out of the circuit. (a) What is the initial current in the resistor, just after S2is closed and S1is opened? (b) What is the current in the resistor att=4×10-4s? (c) What is the potential difference between points b and c at t=4×10-4s? Which point is at a higher potential? (d) How long does it take the current to decrease to half its initial value?

Short Answer

Expert verified

a)i=0.25Ab)i=0.137Ac)V=32.9Vd)t=0.462ms

Step by step solution

01

Calculate the initial current

The formula to find the current in an LR circuit is i=I0e-tRL.

To find the initial current, put t=0

localid="1664256507202" i=I0e-tRLi=I0e0i=I0i=εRi=60240i=0.25A

02

Find the current at the given time

In the formula of current, put t=4×10-4s.

i=I0e-tRLi=0.25e-4×10-4×2400.16i=0.137A

03

Find the potential difference by multiplying current and resistance Step 3: Find the potential difference by multiplying current and resistance

V=IRV=0.137×240V=32.9V

04

Use the current equation to find the time for half the value

According to the situation,i=0.5I0.

role="math" localid="1664256476180" i=0.5I0=I0e-tRL0.5=e-tRLt=-LRln0.5t=-0.16240ln0.5t=0.462ms

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