The sinusoidal potential difference V_1 shown | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2 W$

Vedclass · Answer

(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?

Similar Questions

$A$ current in a circuit is given by $i = 3 + 4 \sin \omega t$. The effective value of the current is

Two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltages? (Frequency in $Hz$, Phase lead of $N$ over $M$ in radians)

The phase difference between the voltage and the current in an $AC$ circuit is $\pi / 4$. If the frequency is $50 \ Hz$,then this phase difference will be equivalent to a time of:

If an $AC$ main supply is given to be $220\,V$,the average $emf$ during a positive half cycle will be.....$V$.

The instantaneous value of current in an $A.C.$ circuit is $I = 3 \sin \left(50 \pi t + \frac{\pi}{4}\right) \text{ A}$. The current will be maximum for the first time at

Vedclass Test Series

Exam Paper Generator

Online Exam Module