Question

A sinusoidal voltage $v=200\sin 314t$ is applied to a resistor of $10\Omega$ resistance. Calculate (i) rms value of the voltage (ii) rms value of the current (iii) power dissipated as heat in watt

Answer 1

Step 1. Given data:

Sinusoidal voltage $v=200\sin 314t$

Resistance of the resistor $\left(R\right)$ = $10\Omega$

Step 2. Formula used:

RMS voltage, $V_{rms}=\frac{V_m}{\sqrt{2}}$, where $V_m=$maximum voltage

RMS current, $I_{rms}=\frac{V_{rms}}{R}$ ,

Power dissipated, $P=V_{rms}{I}_{rms}\cos \theta$,

Where $\cos \theta$ is the power factor which is equal to $1$ for pure resistive circuit since voltage and current are in the same phase and the other variables have their usual meanings,

Thus, for the given case,

Power dissipated, $P=V_{rms}{I}_{rms}$,

Step 3. Calculations:

We know, the general formula of sinusoidal voltage is $v=V_m.\sin \omega t$, where $\omega =$angular frequency, $t=$time.

Comparing this equation with the given equation, we get

Maximum voltage $V_m$ = $200V$.

(I) RMS ( root mean square) voltage, $V_{rms}=\frac{V_m}{\sqrt{2}}$,

$\Rightarrow$ $V_{rms}=\frac{200}{\sqrt{2}}=141.4V$

(ii) RMS current, $I_{rms}=\frac{V_{rms}}{R}$ ,

Where $I_{rms}=$root mean square current.

Resistance of the resistor $\left(R\right)$ = $10\Omega$

$\Rightarrow$ $I_{rms}=\frac{141.4}{10}=14.14A$

(iii) Power dissipated, $P=V_{rms}{I}_{rms}$,

$\Rightarrow$$P=141.4\times 14.14=1999.69\cong 2000W$

Hence, the calculated values are:
(I) RMS value of voltage, $V_{rms}=141.4V$
(ii) RMS value of current, $I_{rms}=14.14A$
(iiI) Power dissipated $P=2000W$