/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q27E 聽In Fig. 30.11, suppose that 蔚... [FREE SOLUTION] | 91影视

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Chapter 4: Q27E (page 1014)

In Fig. 30.11, suppose that 蔚=60V, R=240惟, 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脳10-4s? (c) What is the potential difference between points b and c at t=4脳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=蔚Ri=60240i=0.25A

02

Find the current at the given time

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

i=I0e-tRLi=0.25e-4脳10-4脳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脳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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