Question

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 5.3Ω and 23.4 W, 41.6Ω and 11.2 W, and 12.7Ω and 10.9 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

Answer 1

Power dissipated by a resistor is given as

P = I*I*R

So the first step is to determine what the maximum allowed current would be for each resistor.

23.4W = I^2*5.3

I^2 = 23.4/5.3 = 4.415

I = square root of (4.415) = 2.10 amps

11.2 = I^2*41.6, I = 0.52 amps

10.9 = I^2*12.7, I = 0.93 amps

Now since all three resistors are in series, the current flowing through one resistor will be the same for all three resistors. The lowest maximum current is for the 41.6 ohm resistor at 0.52 amps. So the voltage must not force more than 0.52 amps otherwise the 41.6 ohm resistor will burn up.

The total resistance of the circuit is the sum of all three resistors assuming an ideal battery.

So R-total = 5.3+41.6+12.7 = 59.6 ohms.

E = I*R

so the maximum battery voltage is E = .52*59.6 = 30.99 Volts

P=V^2/R=(30.99^2)/59.6=16.11 W