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(a) Find the current through the battery and each resistor in the circuit shown in Fig. P26.62. (b) What is the equivalent resistance of the resistor network?

Short Answer

Expert verified

(a) I=10.0AI2=2.0AI1=5.34A


(b) The equivalent resistance is 1.40Ω.

Step by step solution

01

Concept Introduction

Kirchoff’s law state that at any junction the algebraic sum of all currents should be always equal to zero.

The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero

02

About circuits and their equations.

Consider the following circuit, where

We need to find the current through each one of the resistors and in the battery(by using the junction rule we will find the current). Apply the loop rule to the left loop(clockwise) we get

ε-2IR1+3I2R1=0orε-I1(R1+R2)+I2R2=0

Apply the loop rule to the right upper loop(clockwise), and we get,

I1R1-(I-I1)R2+I2R2=0I1(R1+R2)-IR2+I2R1=0

And to the right bottom loop( also clockwise), we get,

-I2R1-(I-I1+I2)R1+(I1-I2)R2=0I1(R1+R2)-IR1+I2(2R1+R2)=0

To simplify the problem, we will use the substitution R2=2R1, and we get

ε3I1R1+I2R1=03l1R12R1+I2R1=03l1R1R1+4I2R2=0

We solve this equation for I,I2,andI1the first we subtract the third equation in (4) from the second equation to eliminate

-IR1+5I2R1=0I=5I2

03

Calculations 

And the second equation in (4) to the first one, so we get,

ε-2IR1+3I2R1=0

Substitute from (5)

ε-10I2R1+3I2R1=0

Now, we will have,

l2=£7R1

Substitution of the given data, we get,

I2=ε7R1=14.0V7×1.0Ω=2.0A

Substitute into (5) we get,

I=5l2=5×(2.00A)=10.0A

Which is the current through the battery, substitute into the first equation into (4), so we get,

ε3I1R1+I2R1=0I1=14.0V3×1.0Ω+2.0A31=ε3R1+I23I1=5.34A

Thus,

I=10.0AI2=2.0AI1=5.34A

using these currents you get any current in the circuit as shown in the figure.

(b) The equivalent resistance equals the potential difference across the resistor network, divided by the current through the network which is I so,

Req=ε1=14.0V10.0A=1.40Ω

Hence the equivalent resistance is 1.40Ω

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