Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 116P

(a)

To determine

Find the current through the ammeter A1.

(a)

Expert Solution
Check Mark

Answer to Problem 116P

The current through the ammeter A1 is 1.91 A.

Explanation of Solution

Write the equation for current.

I=VReq (I)

Here, I is the current, Req is the equivalent resistance and V is potential difference.

Consider left top 2.00 Ω resistor as R1, left bottom 2.00 Ω resistor as R2, middle 2.00 Ω resistor as R3, 3.00 Ω resistor as R4, 6.00 Ω resistor as R5, resistance of ammeter A1 as R6 and resistance of ammeter A2 as R7

For ammeter A1, the resistors R4 and R5 are in parallel then the equivalent resistance is,

1R'=1R4+1R5R'=(1R4+1R5)1

This resistance is in series with R7, then the equivalent resistance is,

R''=R7+R'

This resistance is parallel with R3, then the equivalent resistance is,

1R'''=1R3+1R''R'''=11R3+1R''

This resistance is in series with R1, R6 and R2, then the final equivalent resistance is,

Req=R1+R6+R'''+R2

Substitute the value of R', R'' and R''' in the above equation.

Req=R1+R6+11R3+1R7+(1R4+1R5)1+R2 (II)

Conclusion:

Substitute 2.00 Ω for R1, 2.00 Ω for R2, 2.00 Ω for R3, 3.00 Ω for R4, 6.00 Ω for R5, 0.200 Ω for R6 and 0.200 Ω for R7 in equation II.

Req=2.00 Ω+0.200 Ω+112.00 Ω+10.200 Ω+(13.00 Ω+16.00 Ω)1+2.00 Ω=4.20 Ω+1.048 Ω=5.248 Ω

Substitute 5.248 Ω for Req and 10.0 V for V in equation I.

I=10.0 V5.248 Ω=1.905 A1.91 A

Therefore, the current through the ammeter A1 is 1.91 A.

(b)

To determine

Find the current through the ammeter A2.

(b)

Expert Solution
Check Mark

Answer to Problem 116P

The current through the ammeter A2 is 0.908 A.

Explanation of Solution

Consider left top 2.00 Ω resistor as R1, left bottom 2.00 Ω resistor as R2, middle 2.00 Ω resistor as R3, 3.00 Ω resistor as R4, 6.00 Ω resistor as R5, resistance of ammeter A1 as R6 and resistance of ammeter A2 as R7

For ammeter A2, the resistors R4 and R5 are in parallel then the equivalent resistance is,

1R'=1R4+1R5R'=(1R4+1R5)1

This resistor is in series with R7, then the effective resistance is,

Ref=R7+R'

Substitute the value of R' in the above equation.

Ref=R7+(1R4+1R5)1

Conclusion:

Substitute 2.00 Ω for R1, 2.00 Ω for R2, 2.00 Ω for R3, 3.00 Ω for R4 6.00 Ω for R5, 0.200 Ω for R6 and 0.200 Ω for R7 in above equation.

Ref=0.200 Ω+(13.00 Ω+16.00 Ω)1=2.20 Ω

From the previous part the effective resistance right of ammeter A1 is 1.048 Ω. Then the potential difference across this resistance is,

(1.905 A)(1.048 Ω)=1.997 V

This is the voltage across the ammeter A2.

Then current through ammeter A2 is 1.997 V2.20 Ω=0.908 A

Therefore, the current through the ammeter A2 is 0.908 A.

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