Three resistors having resistances of R 1 = 1.57? , R 2 = 2.37? and R 3 = 4.65?

Question

Three resistors having resistances of R1 = 1.57? , R2 = 2.37? and R3 = 4.65? respectively, are connected in series to a 27.6V battery that has negligible internal resistance.

1)Find the equivalent resistance of the combination.

2)Find the current in each resistor.

3)Find the total current through the battery.

4)Find the voltage across each resistor.

5)Find the power dissipated in each resistor.

6)Which resistor dissipates the most power, the one with the greatest resistance or the one with the least resistance? Explain why this should be.

Explanation / Answer

a)

equivalent resistance

Req=1.57+2.37+4.65

Req=8.59 ohms

b)

Total Current flowing in the circuit

I=V/Req=27.6/8.59

I=3.21 A

In series same current will flow through all resistors,Current flowing through each resistor is

I1=3.21 A

I2=3.21 A

I3=3.21 A

c)

Total current through the battery

I=3.21 A

d)

Voltage across each resistor

V=IR

V1=3.21*1.57=5.04 Volts

V2=3.21*2.37=7.61 Volts

V3=3.21*4.65=14.93 Volts

e)

Power dissiated in each resistor is

P=I2R

P1=(3.21)2*1.57=16.18 Watts

P2=(3.21)2*2.37=24.42 Watts

P3=(3.21)2*4.65=47.92 Watts

f)

Greatest resistance ,since power dissipated is directly proprtional to resistance

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