Question

Interactive Solution 20.47 provides one approach to problems like this one. Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.2Ω and 22.7 W, 33.8Ω and 12.7 W, and 14.7Ω and 10.1 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)?

*** The answers are not 20.5, 7.70, 44.766,45.5,10.1, 23.5** these have already been given to me and are incorrect

Answer 1

a)

total resistance of the circuit = 6.2 + 33.8 + 14.7 = 54.7 ohm

Let the voltage of the battery be V

So, total current through the circuit, I = V/54.7

So, power consumed by the 1st resistance = I^2*R1

= (V/54.7)^2*6.2

Now for rated wattage, (V/54.7)^2*6.2 = 22.7

So, V = 104.7 V

Similalrly, for the 2nd resistor,

(V/54.7)^2*12.7 = 33.8

So, V = 89.2 V

and for the 3rd resistor,

(V/54.7)^2*10.1 = 14.7

So, V = 66 V

So, greatest voltage = 66 V <------- lowest value of voltage will give the greatest voltage the battery can have

b)

Power delivered, P = V^2/Rnet

= 66^2/(54.7)

= 79.6 W