Chapter 18, Problem 21P

### College Physics

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

Chapter
Section

### 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.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 Îµ is 250â€‰V and resistance R is 1.00â€‰kÎ© .

Explanation:

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

From the above figure,

• Îµ 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 ,Â the current passing from c to e is I2 ,

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

Apply Kirchhoffâ€™s voltage law to the left and right loop.

From the left loop abca,

Îµâˆ’R(I1âˆ’I)âˆ’4RI1=0Îµ+RI=RI1+4RI1I1=Îµ+RI5R=15(ÎµR+I)

From the right loop cedc,

âˆ’3RI2+2Îµâˆ’2R(I2+I)=05RI2=2Îµâˆ’2RII2=2(Îµâˆ’RI)5R=25(ÎµRâˆ’I)

Conclusion:

Formula to calculate current by using Kirchhoffâ€™s junction rule, to caec

âˆ’4RI1+3RI2=04RI1=3RI2

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

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