Three resistors having resistances of R 1 = 1.52 , R 2 = 2.32 and R 3 = 4.94 res

Question

Three resistors having resistances of R1 = 1.52 , R2 = 2.32 and R3 = 4.94 respectively, are connected in series to a 28.4 V battery that has negligible internal resistance.

Find the equivalent resistance of the combination.

Find the current in each resistor.

Find the total current through the battery.

Find the voltage across each resistor.

Find the power dissipated in each resistor.

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 is given as

Req = R1 + R2 + R3 = 1.52 + 2.32 + 4.94 = 8.78 ohm

b)

In series , current flows same through each resistor . Current flowing is given as

i = V/Req = 28.4/8.78 = 3.2 A

c)

itotal = total current = i = 3.2 A

d)

V1 = Voltage across R1 = i R1 = (3.2) (1.52) = 4.86 Volts

V2 = Voltage across R2 = i R2 = (3.2) (2.32) = 7.42 Volts

V3 = Voltage across R3 = i R3 = (3.2) (4.94) = 15.8 Volts

e)

P1 = power dissipated in R1 = V1 i = 4.86 x 3.2 = 15.6 Watt

P2 = power dissipated in R2 = V2 i = 7.42 x 3.2 = 23.7 Watt

P3 = power dissipated in R3 = V3 i = 15.8 x 3.2 = 50.6 Watt

The one greatest resistance dissipates greatest energy.

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