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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

Taking R = 1.00 kΩ and ε = 250 V in Figure P18.21, determine the direction and magnitude of the current in the horizontal wire between a and e.

images

Figure P18.21

To determine
The magnitude of the current and the direction in the horizontal wire between a and e.

Explanation

Given Info: Voltage ε is 250V and resistance R is 1.00 .

Explanation:

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

From the above figure,

  • ε is voltage in the circuits
  • R is corresponding resistance in the circuits
  • I,I1,I2 is corresponding current in the circuits

Consider, the current passing from c to a is I1 , the current passing from c to e is I2 ,

the current passing from a to e is I voltage is ε and resistance is R

Apply Kirchhoff’s voltage law to the left and right loop.

From the left loop abca,

εR(I1I)4RI1=0ε+RI=RI1+4RI1I1=ε+RI5R=15(εR+I)

From the right loop cedc,

3RI2+2ε2R(I2+I)=05RI2=2ε2RII2=2(εRI)5R=25(εRI)

Conclusion:

Formula to calculate current by using Kirchhoff’s junction rule, to caec

4RI1+3RI2=04RI1=3RI2

Use 1/5[(ε/R)+I] for I1 and (2/5)[(ε/R)I] for I2 in the above equation to find expression for I

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