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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

Find the potential difference across each resistor in Figure P18.31.

images

Figure P18.31

To determine
The potential difference across in each resistor.

Explanation

Given Info:

Resistance R1 is 5.00Ω , Resistance R2 is 4.00Ω , Resistance R3 is 3.00Ω , Resistance R4 is 2.00Ω , The expression for I2 is 3.00V0.800ΩI3 , the expression for I1 is 1.80V0.800ΩI3 , the current in the circuits I1 is 0.324A , I3 is 1.846A , I2 is 1.524A , Resistance R1 is 5.00Ω , Resistance R2 is 4.00Ω , Resistance R3 is 3.00Ω , Resistance R4 is 2.00Ω .

Explanation:

Voltage ε1 is 12.0V , Voltage ε2 is 3.00V , Voltage ε3 is 18.0V

The below figure shows the direction of current flow in the circuit.

  • ε1,ε2,ε3 is voltage in the circuits
  • R1,R2,R3,R4 is corresponding resistance in the circuits
  • I1,I2,I3 is corresponding current in the circuits

Formula to calculate the expression for current by Kirchhoff’s voltage law

From the Left loop

ε1ε2R1I1R2I3=0

Substitute 5.00Ω for R1 , 4.00Ω for R2 , 3.00V for ε2 12.0V for ε1 in the above equation to find expression for I1

12.0V3.00V5.00ΩI14.00ΩI3=09.00V5.00ΩI14.00ΩI3=05.00ΩI1=9.00V4.00ΩI3I1=1.80V0.800ΩI3 (I)

From the right loop

ε3ε2R2I3(R3+R4)I2=0

Substitute 4.00Ω for R2 , 3.00Ω for R3 , 2.00Ω for R4 , 12.0V for ε2 , 18.0V for ε3 in the above equation to find expression for I1

18.0V3.00V4.00ΩI3(3.00Ω+2.00Ω)I2=015.0V4.00ΩI25.00ΩI2=05.00ΩI2=15.0V4.00ΩI3I2=3.00V0.800ΩI3 (II)

Conclusion:

Thus, from the Left loop the expression I1 is 1.80V0.800ΩI3 and from the right loop the expression I2 is 3

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