Three resistors having resistances of R 1 = 1.67 , R 2 = 2.39 and R 3 = 4.82 res

Question

Three resistors having resistances of R1 = 1.67 , R2 = 2.39 and R3 = 4.82 respectively, are connected in series to a 27.5 V battery that has negligible internal resistance.

Part A

Find the equivalent resistance of the combination.

Part B

Find the current in each resistor.

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

Part C

Find the total current through the battery.

Part D

Find the voltage across each resistor.

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

Part E

Find the power dissipated in each resistor.

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

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).

Three resistors having resistances of R1 = 1.67 , R2 = 2.39 and R3 = 4.82 respectively, are connected in series to a 27.5 V battery that has negligible internal resistance.

Part A

Find the equivalent resistance of the combination.

Req = ??

Part B

Find the current in each resistor.

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

IR1,IR2,IR3 = ??? A

Part C

Find the total current through the battery.

I = ??? A

Part D

Find the voltage across each resistor.

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

VR1,VR2,VR3 = ??? V

Part E

Find the power dissipated in each resistor.

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

PR1,PR2,PR3 = ??? W

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).

Explanation / Answer

Part A)

for the resistances in series ,

Req = R1 + R2 + R3

R = 1.67 + 2.39 + 4.82 = 8.88 ohm

equivalent resistance is 8.88 Ohm

voltage = V = 27.5 V

part B)

according to ohms law,

I = V/r = 27.5 / 8.88 = 3.09 A

as current remain same in each resistor when they are in series so R1,R2,R3 will have 3.09A

part c)

as the current is same in all resistances and battery.

the total current through the battery is 3.09 A

Part D)

Using Ohm's law

as V = I*R

VR1 = R1 *I

VR1 = 1.67 * 3.09

VR1 = 5.16 V

--------------------

VR2 = 2.39 * 3.09

VR2 = 7.39 V
--------------

VR3 = 4.82 * 3.09

VR3 = 14.9 V

part E)

as the power in resistance is given as

Power = V * I

PR1 = 5.16 * 3.09 = 15.94 W

PR2 = 7.39 * 3.09 = 22.8 W

PR3 = 14.9 * 3.09 = 46.04 W

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