Three resistors having resistances of R1 = 1.48? , R2 = 2.39? and R3 = 4.81? res

Question

Three resistors having resistances of R1 = 1.48? , R2 = 2.39? and R3 = 4.81? respectively, are connected in series to a 27.5V battery that has negligible internal resistance.

Part A

Find the equivalent resistance of the combination.

Part B

Find the current in each resistor.

IR1,IR2,IR3 =

Part C

Find the total current through the battery.

Part D

Find the voltage across each resistor.

Part E

Find the power dissipated in each resistor.

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.

Explanation / Answer

Net resistance = 8.68 ohm

Current in each resistor is same =27.5/8.68= 3.168 A

CUrrent through the battery = 3.168 A

Volatage across each resistor =
R1 = 4.68 V

R2 =7.57 V

R3 =15.23 V

Power dissipated =V2/R
R1 = 14.79 W

R2= 234.97W

R3 = 48.22 W (ans)

The one with greatest resistance dissipates more power due to more volatage drop.

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