Question

A sinusoidal voltage $\mathrm{V}=200 \sin 314 \mathrm{t}$ is applied to a $10 \Omega$ resistor. Find:
(i) rms voltage,
(ii) rms current and
(iii) power dissipated as heat.

Answer 1

We know that rms voltage is defined as the voltage of a sinusoidal source in case of any electromotive force which is used as the source. The value of the current that is carried by the sinusoidal wave of a source is defined as the rms current. And lastly, we know that when there is production of heat or undesirable quantity of energy which results in the loss of power, then it is known as power dissipation.

Complete step by step answer:
We know that according to the question $\mathrm{V}=200 \sin 314 \mathrm{t}$ and $\mathrm{R}=10 \Omega$ comparing it with original equation we get $, \mathrm{V}_{\mathrm{m}}=200 \mathrm{~V}$.
Now let us solve to find the rms voltage:
The expression to find the rms voltage is given as:
$\Rightarrow \mathrm{V}_{\mathrm{rms}}=\dfrac{\mathrm{V}_{\mathrm{m}}}{\sqrt{2}}$
Now we have to put the values in the above expression to get the answer as:
$\Rightarrow 0.707 \times 200=141.4 \mathrm{~V}$
Hence, the value of rms voltage is 141.4 V
Now we have to find the value of rms current
The expression to find the rms current is given as:
$\Rightarrow \mathrm{I}_{\mathrm{rms}}=\dfrac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{R}}$
Now we have to put the values in the above expression to get the answer as:
$\Rightarrow \dfrac{141.4}{10}=14.14 \mathrm{~A}$
Hence, the value of rms current is 14.14 A.
Now we have to find the value of power dissipation.
The expression to find the power dissipate is given as:
$\Rightarrow \mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \theta$
Now we can say that here $\cos \theta$ is power factor which is equal to 1 for pure resistive circuit since voltage and current are in same phase,
$\Rightarrow \mathrm{P}=141.4 \times 14.14=1999.69 \simeq 2000 \mathrm{~W}$

Hence, the power dissipate is given as 2000 W.

Note: We should know that the effect of power dissipation is very important for us to know. The greater will be the value of the power the more amount of heat will be dissipated by that particular component. This will result in the increase in temperature of that component.