Three resistors having resistances of R1 = 1.85 , R2 = 2.41 and R3 = 4.93 respec

Question

Three resistors having resistances of R1 = 1.85 , R2 = 2.41 and R3 = 4.93 respectively, are connected in series to a 28.1 V battery that has negligible internal resistance.

Part A

Find the equivalent resistance of the combination.

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Part B

Find the current in each resistor.

Answer in the order indicated. Separate your answers using commas.

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Part C

Find the total current through the battery.

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Part D

Find the voltage across each resistor.

Answer in the order indicated. Separate your answers using commas.

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Part E

Find the power dissipated in each resistor.

Answer in the order indicated. Separate your answers using commas.

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Part F

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

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

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Explanation / Answer

The effective resistance of the series combination of 3 resistors can be written as follows.

R = R1+R2+R3

R = 1.85 + 2.41 + 4.93

R = 9.19

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In series combination, the current through each resistor is same,

I = V / R = 28.1 / 9.19 = 3.057 A.

Round off the result to 3 significant digits

The current through each resistor would be I = 3.06 A.

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Total current in the circuit is same, so the current through the battery is

I = 3.06 A.

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Voltage across R1 = IR1 = ( 3.06 ) ( 1.85 ) = 5.66 V

Voltage across R2 = IR2 =  ( 3.06 ) ( 2.41 ) = 7.37 V

Voltage across R3 = IR3 =  ( 3.06 ) ( 4.93 ) = 15.1 V

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