Question

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 5.4Ω and 28.3 W, 33.8Ω and 9.33 W, and 15.1Ω and 12.6 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

here,

a)

for R1 = 5.4 ohm

P1 = 28.3 ohm

let the maximum current be I1

P1 = I1^2 * R1

on substituting values

I1 = 2.289 A

for R2 = 33.8 ohm

P2 = 9.33 ohm

let the maximum current be I2

P2 = I2^2 * R2

on substituting values

I2 = 0.5253 A

for R3 = 15.1 ohm

P3 = 12.6 ohm

let the maximum current be I3

P3 = I3^2 * R3

on substituting values

I3 = 0.91 A

as the minimum amount of safe current is for I2 = 0.5253 A

so, the greatest voltage that the battery can have without one of the resistors burning up , V = I2 * ( R1 + R2 + R3)

V = 0.5253 * ( 5.4 + 33.8 + 15.1) V = 28.5 V

b)

the power delivered by the battery to the circuit , P = V * I2

P = 28.5 * 0.5253 W = 14.99 W