Question: The total power in a parallel circuit with four resistors of equal value is 570W. If one of the resistors becomes open circuited, what is the power dissipated by each of the remaining resistors?
My Attempt:
\$P_1 + P_2 + P_3 + P_4 = 570\$
\$P_1 + P_2 + P_3 + \frac{v^2}{R} = 570\$
Open circuit causes \$R-> \infty \$
\$P_1 + P_2 + P_3 + \frac{v^2}{\infty} = 570\$
\$P_1 + P_2 + P_3 + 0 = 570\$
\$3P = 570\$
\$P = 190\$
This is not the correct answer, so what am I doing wrong?
This line:
\$P_1 + P_2 + P_3 + \frac{v^2}{\infty} = 570\$
is not valid, because total power changes as there is less loads (resistors) in the circuit.
You can see this by calculating the equivalent resistance, in this case was \$\frac{R}{4}\$ and now is \$\frac{R}{3}\$, so the resistance is higher and decreases the power.
Actually each resistor in parallel is responsible for a part of the total power, so, before, you had:
\$P_1 = P_2 = P_3 = P_4 = \frac{570}{4} = 142,2\$
So as long as voltage remains the same over the resistors, dissipated power will remain the same.
Now you have only three resistors, so:
\$P_{total} = P_1 + P_2 + P_3 = 3*142,2 = 427,5\$
