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Consider the circuit shown in Fig.P26.63. (a) What must the emf Ԑ of the battery be for a current of 2.00 A to flow through the 5.00-V battery as shown? Is the polarity of the battery correct as shown? (b) How long does it take for 60.0 J of thermal energy to be produced in the10.0Ω resistor?

Short Answer

Expert verified

(a)ε=-109Vmust the emf Ԑ of the battery be for a current of 2.00 A to flow through the 5.00-V battery as shown and the polarity of the battery is reversed.

(b) It takes 10 seconds for 60.0 J of thermal energy to be produced in the resistor.

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

Appling general concepts

(a)

Consider the following circuit, whereR1=5.00Ω,R2=10.0Ω,R3=15.0Ω,R4=20.0Ω,

R5=30.0Ω,R6=60.0Ω,ε1=10.0V

we need to find the emfand such that the current flowing through the 5.00-V battery is I = 2.00 A.

We can simplify the circuit by finding the equivalent resistance and the equivalent emfas shown in the second figure.

We know that the resistance adds in series and its reciprocal add in parallel, so we have,

R'=2R1+Rparallel (1)

where,

1Rparallel=1R1+2R6+1R2+R4

we can express all the resistance in terms of R1, that is R2=R1, R3=3R1,R4=4R1,R5=6R1,R6=12R1and thus

1Rparallel=16R1+212R1+16R1=12R1Rparallel=2R1

substitute into(1) we get,

R=2R1+2R1=4R1=20.0Ω (2)

also, we have,

ε'=ε2-ε1=5.00V (3)

assume that the current flowing through the resistor R3in the second figure is l1, and the current flowing in the resistor R4, is l2, from the junction rule we can write,

I=I1+I2 (4)

03

Calculation of EMF

Now we can apply the loop rule two times first by passing through the branch where the resistor R3exists and second by passing through the branch where the resistor localid="1668334078768" R4exists (counterclockwise), so we get

ε'+IR'+I1R3+ε=0 (5)

and,

ε'+IR'+I2R4=0 (6)

now we have three equations(4), (5), and (6), we need to solve them for ε, first, subtract (6) from (5), we get,

I1R3+ε=I2R4

or,

l2=l1R3R4+εR4

but R3=3R1,andR4=4R1, therefore,

I2=3I14+ε4R1

substituting into (4) we get,

I=7I14+ε4R1I1=47I+ε7R1

substitute (5) to get,

ε+IR+47IR3εR37R1+ε=0ε+IR+127IR13ε7+ε=0ε=47ε+IR+127R1

substitute with the given to get,

ε=7105.00V+(2.00A)(20.0Ω)+127(2.00Ω(5.00Ω))ε=109V

04

Calculation of time

Since the sign of the emf is negative, then the polarity of the battery is not correct and it must be reversed

(b) Now we need to find how long it takes for 6.00 J of thermal energy to be produced in the 10.0Ωresistor, to do that we need to find the current flowing through this resistor, the current flowing through the parallel resistors network is I= 2.00 A, and the potential across each one of the resistors is the same, which equals the current flowing the parallel resistors network multiplied by their resistance, Rparallelthat is

V=IRparallel=2IR1

the current flowing through the resistor R₂equals this potential divided by the resistance of its branch, that is.

I2=VR2+R4=2R1R2+R4=2R12R1+4R1=13

the energy of this resistor equals its power multiplied by the time, that is

ε2=I22R2t=l2R1t9

solve for t to get

t=9ε22l2R1

substitute with I=2.00A,R1=5.00Ω,and ε2=60.0J

t=9(60.0J)2(2.00A)2(5.00Ω)t=13.5s

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