Question

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.7Ω and 15.9 W, 30.4Ω and 9.12 W, and 16.3Ω and 12.3 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

a) greatest voltage = 29.25 V

b) power = 16 W

Step-by-step explanation:

The total resistance R of the three resistors in series is:

`R = (6.7 + 30.4 + 16.3) \Omega = 53.4 \Omega`

a) The greatest current I is the one that will burn the resistor with lower power rating, which is 9.12 W:

`P_(max) = I_(max)^2 R = I_(max)^2 30.4\Omega = 9.12W\\I_(max) = 0.54 A`

The voltage is:

`V_(max)=IR = 0.54*53.4V= 29.25 V`

b) When the current is 0.54 A, the power is:

`P = RI^2=53.4*0.3 W = 16W`