Three resistors having resistances of R1 = 1.53 ? , R2 = 2.37 ? , and R3 = 4.73

Question

Three resistors having resistances of R1 = 1.53?, R2 = 2.37?, and R3 = 4.73?respectively, are connected in parallel to a 28.3 V battery that has negligible internal resistance.

a) Find the equivalent resistance of the combination.

b) Find the current in each resistor.

c) Find the total current through the battery.

d) Find the voltage across each resistor.

e) Find the power dissipated in each resistor.

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

Effective resistance is 1/R = 1/R1 + 1/R2 +1/R3

                             R = 0.78

As it is parallel connection, Potential difference is same across all resistors.

I = V/R

Current in R1 = 18.5 A

Current in R2 = 11.9 A

Current in R3 = 6 A

Total current = 28.3/0.78 = 36.3 A

Voltage across each resistor is 28.3 V.

Power dissipiaated in resistor = p = I2R

Power dissipiaated in resistor R1 = 523.6 J

Power dissipiaated in resistor R2 = 336.8 J

Power dissipiaated in resistor R3 = 169.8 J

The one with least resistance dissipiated more power. Because the circuit is parallel, more current lends to flow through least resistance than higher resistance. Hence more energy is dissipiated in least resistance.

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