Question

The average power dissipated by a resistor connected to a sinusoidal emf is 5.0 W.

a) What is Pavg if the resistance R is doubled?
Pavg=IR2R
Pavg=IR22R ∴=IR2R ∴Pavg=2.5 W
b) What is P_{\rm avg} if the peak emf {\cal{E}}_0 is doubled?
c) What is P_{\rm avg} if both are doubled simultaneously?

Answer 1

The average power if both the resistance and peak emf are doubled simultaneously is 10 W.

a)

The average power is:

$P_{avg}=\frac{(V_{RMS})^2}{R}$

If the resistance is doubled. The new average power is:

$P_{avg}'=\frac{(V_{RMS})^2}{2R}=0.5*\frac{(V_{RMS})^2}{R} =0.5*5=2.5\ W$

b)

$P_{avg}=\frac{(V_{RMS})^2}{R}= \frac{(E_o/\sqrt{2} )^2}{R}$

If the peak emf is doubled:

$P_{avg}'=\frac{(2E_o/\sqrt{2} )^2}{R}=4*\frac{(E_o/\sqrt{2} )^2}{R}=4*5=20\ W$

c)

$P_{avg}=\frac{(V_{RMS})^2}{R}= \frac{(E_o/\sqrt{2} )^2}{R}$

If both are doubled simultaneously:

$P_{avg}'=\frac{(2E_o/\sqrt{2} )^2}{2R}=2*\frac{(E_o/\sqrt{2} )^2}{R}=2*5=10\ W$

The average power if both the resistance and peak emf are doubled simultaneously is 10 W.

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