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Chapter 26 Part 1--Examples

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Problem

If a ohmmeter is placed between points a and b in the circuits to right, what will it read?

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Circuit a

The 75 and 40 resistors are in parallel

The 25 and 50 resistors are in parallel

R75-40=26.09 R50-25=16.67 Their total is 42.76

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Circuit a—Cont’d

The circled network is in parallel with the 50 resistor so their combined resistance is 23.05 .

This resistor is in parallel with the original 100 resistor so the total resistance is 18.7

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Circuit b

The 60 and 20 resistor are in parallel

The 20 is in series with the 30 and 40 parallel network.

R30-40 =18.0 R20-30-40= 38.0 R38-60=23.3

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Circuit b—cont’d

The 23.03 equivalent network is in series to the 7 resistor

This equivalent 30.03 resistor is parallel to the 10 resistor so

Req=7.5

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Problem

In the circuit shown, 1. What must be the

EMF of the battery in order for a current of 2 A to flow through the 5 V battery?

2. How long does it take for 60 J of thermal energy to be produced in the 10 resistor?

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Step 1 Reduce the Resistors

10 + 20 =30 (1/30)+1/60 +1/60

=4/60 so Req=15

1/15+1/30=3/30 so RT=10

10+5+5=20 So 20 in the

upper network

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Step 2 Reduce the EMF

-5 + 10=+5 V So the circuit

becomes:

20 5 V

EMF15

20

2 A

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Using our loop rules

-(2)*20-5-20I1=0 I1=-2.25 A 2=-2.25+I2

I2=4.25 -EMF-4.25*15+20*(-

2.25) EMF=-108.75 V Need to reverse the

battery….

20 5 V

EMF15

20

2 A

I1

I2

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2 Amps into the 10 resistor

Since the equivalent resistance in the upper network is 10 and 2 A runs through it, there is a potential difference of 20 V across each of the legs

10+20=30 so the current is 20/30 A=2/3 A P=i2r so 4/9*10=40/9=4.444 W or J/s

60=4.444 * t t =13.5 s