Respuesta :
Answer;
= 2493 W
Explanation;
The total resistance of 3 resistors in series = R + R + R = 3 * R
The total resistance of 3 resistors in parallel = 1 ÷ (1/R + 1/R + 1/R) = ⅓ R
At the same voltage, V, the total current in the series circuit will be
= V ÷ (3 * R)
= ⅓ * V/R.
At the same voltage, V, the total current in the parallel circuit will be;
= V ÷ (⅓ R)
= 3 * V/R
At the same voltage, the parallel circuit will have 9 times the amperage of the series circuit.
Power = Volts * amps
Power of series circuit = V * ⅓ * V/R = ⅓ * V^2/R
Power of parallel circuit = 3 * V^2/R
But, power of series is 277 W
1/3 (V²/R) = 277
(V²/R) = 277 × 3
Power in parallel = 3 V²/R
= 277 × 3 × 3
= 2493 W
Answer:
2493W
Explanation:
The relationship between the power P consumed by a resistor R and the potential difference V across the it is given by;
[tex]P=\frac{V^2}{R}..............(1)[/tex]
According to the problem stated, the potential difference V is constant.
If V is kept constant, then we can write the following;
[tex]PR=V^2..............(2)[/tex]
Equation (2) implies that we can write the following as long as V is kept constant;
[tex]P_1R_1=P_2R_2=...=P_nR_n...........(3)[/tex]
We can simply write equation (3) as;
[tex]P_1R_1=P_2R_2..............(4)[/tex]
From the problem stated, [tex]R_1[/tex] is a series combination of three identical resistors while [tex]R_2[/tex] is their parallel combination. Let the value of each resistor be R, hence;
[tex]R_1=R+R+R\\R_1=3R[/tex] (series combination)
[tex]\frac{1}{R_2}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\\frac{1}{R_2}=\frac{3}{R}\\hence;\\R_2=\frac{1}{3}R[/tex] (parallel combination)
Given;
[tex]P_1=277W[/tex] for the series combination, are supposed to find [tex]P_2[/tex] for the parallel combination. Hence we make the necessary substitutions into equation (4) as follows;
[tex]277*3R=P_2*\frac{1}{3}R\\831R=P_2*\frac{1}{3}R[/tex]
R cancels out from both sides and we get the following,
[tex]831=\frac{P_2}{3}\\hence;\\P_2=831*3\\P_2=2493W[/tex]
