The power, in watts, dissipated as heat in a resistor varies jointly with the resistance, in ohms, and the square of the current, in amperes.A 30-ohm resistor carrying a current of 1 ampere dissipates 30 watts.
How much power is dissipated in a 50-ohm resistor carrying a current of 8 amperes?
A) $\quad P = 50$ Watts
B) $\quad P = 8$ Watts
C) $\quad P = 3,200$ Watts
D) $\quad P = 5$ Watts
E) $\quad P = 40$ Watts

Question

The power, in watts, dissipated as heat in a resistor varies jointly with the resistance, in ohms, and the square of the current, in amperes.A 30-ohm resistor carrying a current of 1 ampere dissipates 30 watts.
How much power is dissipated in a 50-ohm resistor carrying a current of 8 amperes? 
A) $\quad P = 50$ Watts
B) $\quad P = 8$ Watts
C) $\quad P = 3,200$ Watts
D) $\quad P = 5$ Watts
E) $\quad P = 40$ Watts

Quizplus · Accepted Answer

We know that power (P) varies jointly with resistance (R) and the square of the current (I). So we can write the relation as:

P = k * R * I^2

where k is the constant of variation.

To find the value of k, we can use the given information:

"A 30-ohm resistor carrying a current of 1 ampere dissipates 30 watts"

Substituting these values in the equation:

30 = k * 30 * 1^2

Simplifying, we get:

k = 1

Now we can use this value of k to find the power dissipated in the second resistor:

P = 1 * 50 * 8^2 = 3,200 watts

Therefore, the answer is C.

The Power, in Watts, Dissipated as Heat in a Resistor P=50\quad P = 50P=50

The Power, in Watts, Dissipated as Heat in a Resistor $\quad P = 50$