Three resistors having resistances of R 1 = 1.72 , R 2 = 2.51 and R 3 = 4.73 res

Question

Three resistors having resistances of R1 = 1.72 , R2 = 2.51 and R3 = 4.73 respectively, are connected in series to a 28.5 V battery that has negligible internal resistance.

Part A:Find the equivalent resistance of the combination.

Part B: Find the current in each resistor.

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

A) Req = R1+R2+R3 = 1.72+2.51+4.73 = 8.96 ohm

B) i = V/Req = 28.5/8.96 = 3.18 A

C) i = 3.18 A

D) V1 = i*R1 = 3.18*1.72 = 5.4696 V
V2 = i*R2 = 3.18*2.51 = 7.9818 V

V3 = i*R3 = 3.18*4.73 = 15.0414 V

E) P1 = i^2*R! = 3.18^2*1.72 = 17.4 W

P2 = i^2*R2 = 3.18^2*2.51 = 25.4 W

P3 = i^2*R3 = 3.18^2*4.73 = 47.8 W

F) Since power dissipation is proportional to the resistance,then greatest resistance dissipates more power

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