Answer:
The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W
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
[tex]P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)[/tex]
If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;
[tex]\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\[/tex]
[tex]P_{parallel} = 405 \ W[/tex]
Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W
