Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 73P

(a)

To determine

Find the equivalent resistance.

(a)

Expert Solution
Check Mark

Answer to Problem 73P

The equivalent resistance is 35.0 Ω.

Explanation of Solution

The sum of individual resistance connected in series and reciprocal of resistance connected in parallel will give the equivalent resistance.

From the given circuit, the resistor R2, R3  and R4  are in parallel then the equivalent resistance is,

1R'=1R2+1R3+1R4R'=(1R2+1R3+1R4)1

This resistance is in series with the R1 and R5 , then the final equivalent resistance is,

Req=R1+R'+R5

Substitute the value of R' in the above equation.

Req=R1+(1R2+1R3+1R4)1+R5

Conclusion:

Substitute, 15.0 Ω for R1 , 40.0 Ω for R2 , 20.0 Ω for R3 , 40.0 Ω for R4 ,and 10.0 Ω for R5 in the above equation.

Req=15.0 Ω+(140.0 Ω+120.0 Ω+140.0 Ω)1+10.0 Ω=35.0 Ω

Therefore, the equivalent resistance is 35.0 Ω.

(b)

To determine

Find the current flows through R1 .

(b)

Expert Solution
Check Mark

Answer to Problem 73P

The current flows through R1  is 0.686 A.

Explanation of Solution

The emf is 24 V.

Write the equation for current.

I=εReq (I)

Here, Req is the equivalent resistance, ε is the emf and I is the current.

Conclusion:

Substitute 24 V for ε and 35.0 Ω for Req in equation I.

I=24 V35.0 Ω=0.686 A

Therefore, the current flows through R1  is 0.686 A.

(c)

To determine

Find the power dissipated in in the circuit.

(c)

Expert Solution
Check Mark

Answer to Problem 73P

The power dissipated in the circuit is 16.5 W.

Explanation of Solution

Write the equation for power dissipated.

P=V2Req (II)

Here, P is the power dissipated and V is the potential difference.

Conclusion:

Substitute 24.0 V for V and 35.0 Ω for Req in equation II.

P=(24.0 V)235.0 Ω=16.5 W

Therefore, the power dissipated in the circuit is 16.5 W.

(d)

To determine

Find the potential difference across R3.

(d)

Expert Solution
Check Mark

Answer to Problem 73P

The potential difference across R3 is 6.9 V.

Explanation of Solution

Use the Kirchhoff’s loop rule to the left hand loop to determine the voltage drop.

Consider V3  as the potential difference across R3

Conclusion:

From Kirchhoff’s loop rule,

24.0 V(0.686 A)(15.0 Ω)V3(0.686 A)(10.0 Ω)=0

From the above equation solve for V3

V2=(24.010.296.86) V=6.85V=6.9V

Therefore, the potential difference across R3 is 6.9 V.

(e)

To determine

Find the current flows through R3.

(e)

Expert Solution
Check Mark

Answer to Problem 73P

The current flows through R3 is 0.34 A .

Explanation of Solution

Write the equation for potential difference.

I=VR (III)

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

Conclusion:

Substitute 6.85V for V and 20.0 Ω for R in equation III.

I=6.85 V20.0 Ω=0.34 A

Therefore, the current flows through R3 is 0.34 A.

(f)

To determine

Find the power dissipated in R3.

(f)

Expert Solution
Check Mark

Answer to Problem 73P

The power dissipated in R3 is 2.4 W .

Explanation of Solution

Consider the resistor R3 as Req

Conclusion:

Substitute 6.85V for V and 20.0 Ω for Req in equation II.

P=(6.85V)220.0 Ω=2.4 W

Therefore, the power dissipated in R3 is 2.4 W.

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Chapter 18 Solutions

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