   Chapter 18, Problem 55AP

Chapter
Section
Textbook Problem

The circuit in Figure P18.55 has been connected for several seconds. Find the current (a) in the 4.00-V battery,(b) in the 3.00-Ω resistor,(c)in the 8.00-V battery, and (d)in the 3.00-V battery.(e)Find the charge on the capacitor. To determine

The value of R.

Explanation

Given info: The power dissipated by the resistance R is 20 W.

The circuit is given as,

• I is the current flowing through the circuit.
• I1 is the current flowing through the resistance R.

Applying Kirchhoff’s loop rule in loop 1,

(5.0Ω)I+(30Ω)(II1)=75V7I6I1=15A

Applying Kirchhoff’s loop rule in loop 2,

(30Ω)(II1)=(40Ω+R)I1I=I1(70Ω+R30Ω)

From the above equations,

I1=450A310+7R

Formula to calculate the power dissipated by the resistance R is,

PR=I12R

Substitute 20 W for PR and re-arrange

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