Question

hree identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 45.0 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference

Answer 1

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Step-by-step explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

`P = (V^2)/(R_t) = (V^2)/(3R) \\\\3P = (V^2)/(R) ------equation(i)`

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

`(1)/(R_t) = (1)/(R_1) +(1)/(R_2) + (1)/(R_3) \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = (1)/(R) +(1)/(R) + (1)/(R) \\\\(1)/(R_t) =(3)/(R) \\\\R_t = (R)/(3) \\\\P = (V^2)/(R_t) = (3V^2)/(R) \\\\P_(parallel) = (3V^2)/(R) ---------equation (ii)\\\\From \ equation \ (i), 3P_(series) = (V^2)/(R), Substitute \ this \ into \ equation \ (ii)\\\\P = 3((V^2)/(R) )\\\\P = 3(3P)\\\\P_(parallel) = 9P_(series)\\\\P_(parallel) = 9(45)\\\\`

`P_(parallel) = 405 \ W`

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W