Question

The average power dissipated by a resistor is $4.0W$. What is $P_R$if:

a. The resistance$R$is doubled while $E_0$is held fixed?

b. The peak emf$E_0$is doubled while $R$is held fixed?

c. Both are doubled simultaneously?

Short answer

The $P_R$are (a) $2W$(b) $16W$(c) $8W$.

Step-by-step solution

Part (a) Step 1: Given information

We are given that average power dissipated by a resistor is$4W$and the resistance localid="1649950857686" $R$is doubled.

Part (a) Step 2: Explanation

As we know that-

$P_o=\frac{E^2}{R}$

This is equation (1)

Now from equation power is inversely proportional to resistance so if we doubled the resistance then power becomes half.

$$\begin{gathered} P=\frac{1}{2}{P}_o \\ P=\frac{1}{2}\left(4W\right) \\ P=2W \end{gathered}$$

Part (b) Step 1: Given information

We are given that average power dissipated by a resistor is $4W$and the peak emf $E_0$is doubled.

Part (b) Step 2: Calculation

From equation (1) we can see that power is directly proportional to the square of emf ($E^2$) so if we double the emf then power becomes four times.

$$\begin{gathered} P=\left(2^2\right)P_o \\ P=\left(2^2\right)4W \\ P=16W \end{gathered}$$

Part (c) Step 1: Given information

We are given that average power dissipated by a resistor is $4W$and both $E_0,R$are doubled,

Part (c) Step 2: Calculation

From equation ($1$) we can see that power is directly proportional to the square of emf and inversely proportional to resistance so if we doubled both simultaneously then power becomes doubled.

$$\begin{gathered} P=\frac{{\left(2E\right)}^2}{2R} \\ P=\frac{4}{2}{P}_o \\ P=2\left(4W\right) \\ P=8W \end{gathered}$$