Question

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.6Ω and 17.2 W, 34.6Ω and 12.0 W, and 20.0Ω 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

Resistance of resistor 1 = R₁ = 6.6

Power rating of resistor 1 = P₁ = 17.2 W

Maximum current that can pass through resistor 1 without burning it up = I₁

P₁ = I₁²R₁

17.2 = (I₁²)(6.6)

I₁ = 1.614 A

Resistance of resistor 2 = R₂ = 34.6

Power rating of resistor 2 = P₂ = 12 W

Maximum current that can pass through resistor 2 without burning it up = I₂

P₂ = I₂²R₂

12 = (I₂²)(34.6)

I₂ = 0.589 A

Resistance of resistor 3 = R₃ = 20

Power rating of resistor 3 = P₃ = 12.6 W

Maximum current that can pass through resistor 3 without burning it up = I₃

P₃ = I₃²R₃

12.6 = (I₃²)(20)

I₃ = 0.793 A

The three resistors are connected in series therefore the current flowing through the resistors is same.

For the greatest voltage we should be able to pass the highest possible current.

The maximum possible current through the series connection without one of the resistors burning up is equal to the smallest value of I₁,I₂,I₃

Current through the circuit = I = I₂ = 0.589 A

Resistance of the circuit = R

R = R₁ + R₂ + R₃

R = 6.6 + 34.6 + 20

R = 61.2

Voltage of the circuit = V

V = IR

V = (0.589)(61.2)

V = 36.05 V

Power delivered by the battery = P

P = I²R

P = (0.589)²(61.2)

P = 21.23 W

a) Greatest voltage the battery can have without one of the resistors burning up = 36.05 V

b) Power delivered by the battery in the circuit = 21.23 W