Question

Three resistors having resistances of 1.10 Ω , 2.40 Ω , and 5.00 Ω are connected in parallel to a 28.0 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.

Answer 1

Solution:

Three resistors are R1=1.1 ohm

R2=2.4 ohm

R3=5ohm

Voltage of the battery V = 28V

a) Find the equivalent resistance of the combination.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = (1/1.1) + (1/2.4) + (1/5)

R = = 0.655 ohm

b) Find the current in each resistor.

each resistor has the batteries voltage across it
V=IR

I1 = V/R1 = 28/1.1=25.45 Amp

I2 = V/R2 = 28/2.4=11.66 Amp

I3 = V/R3 = 28/ 5 = 5.6 Amp

c) Find the total current through the battery.

V=IR

I = V/R = 28/0.655 = 42.75 Amp

d) Find the voltage across each resistor.

same voltage across each resistors, and the voltage is equal to that of the battery (ie 28V)

e) Find the power dissipated in each resistor.

P=I*I*R
I=V/R

So P= (V/R)*(V/R)*R= V*V/R

for the individual resistors:-
P1=V*V/R1 =28*28/1.1 =712.7 W

P2=V*V/R2 =28*28 /2.4=326.66W

P2=V*V/R2 =28*28/5=156.8W

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.

The one with the least resistance dissipates the most power. The one with the least resistance has the most current and all have the same voltage.