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 15-ohm resistor carrying a current of 1 ampere dissipates 15 watts. How much power is dissipated in a 5-ohm resistor carrying a current of 3 amperes?

Answer 1

Answer:

The power dissipated is 45W.

Step-by-step explanation:

The power $P$ varies jointly with resistance $R$, and the square of current $I$:

$P = \alpha I^2R$,

where $\alpha$ is the constant of proportionality.

Now we are told that when $R = 15\Omega$ and $I =1A$, $P = 15W$:

$15 = \alpha (1A)^2*15\Omega$

solving for $\alpha$ we get

$\alpha = 1$,

which gives

$P = I^2R$

With the value of $\alpha$ in hand, we find the power dissipated when $R =5\Omega$ and $I = 3 A:$

$P = (3A)^2(5\Omega )$

$\boxed{P =45W}$

Thus, the power dissipated is 45W.