The sinusoidal potential difference V_1 shown | vedclass

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(D) The rate of heat dissipation (power) in a resistor is given by $P = \frac{V_{rms}^2}{R}$.For the sinusoidal potential difference $V_1$ with peak v

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$2 W$

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(D) The rate of heat dissipation (power) in a resistor is given by $P = \frac{V_{rms}^2}{R}$.For the sinusoidal potential difference $V_1$ with peak value $V_0$,the root-mean-square $(RMS)$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}}$.Thus,the rate of heat dissipation is $W = \frac{(V_0 / \sqrt{2})^2}{R} = \frac{V_0^2}{2R}$.For the square wave potential difference $V_2$ with peak value $V_0$,the voltage alternates between $+V_0$ and $-V_0$. The $RMS$ value for this square wave is $V_{rms} = V_0$.Thus,the new rate of heat dissipation is $W' = \frac{V_0^2}{R}$.Comparing the two expressions:$W' = 2 	imes \left( \frac{V_0^2}{2R} ight) = 2W$.Therefore,the rate of heat dissipation becomes $2W$.

The sinusoidal potential difference $V_1$ shown in the figure is applied across a resistor $R$,producing heat at a rate $W$. What is the rate of heat dissipation when the square wave potential difference $V_2$ as shown in the figure is applied across the same resistor?

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