Question

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

Answer 1

Step 1: Given data

The alternative voltage source is $V=200\sin \left(314t\right)$.

The magnitude of voltage is $V_m=200volt.$

The resistance of the resistor is $10\Omega$.

Step 2: RMS voltage, current, and power dissipation

1. The RMS voltage or RMS current is the square root of the time average of the voltage or current squared.
2. The RMS voltage and RMS current is defined by the form, $V_{rms}=\frac{V}{\sqrt{2}}$ $I_{rms}=\frac{I}{\sqrt{2}}$, where, V and I are the magnitudes of voltage and current.
3. The power dissipation in a circuit as heat is $P_d=V_{rms}{I}_{rms}\cos \theta$, where, $\cos \theta$ is the power factor and for the purely resistive circuit its magnitude is 1.

Step 3: Finding the RMS voltage

As we know, $V_{rms}=\frac{V_m}{\sqrt{2}}$. So,

$$\begin{gathered} V_{rms}=\frac{V_m}{\sqrt{2}}=\frac{200}{\sqrt{2}}=100\sqrt{2} \\ or{V}_{rms}=141.42volt. \end{gathered}$$

Therefore, the RMS voltage is $141.42volt.$

Step 4: Finding the RMS current

As we know, $I_{rms}=\frac{I}{\sqrt{2}}$ . So,

$$\begin{gathered} I_{rms}=\frac{I}{\sqrt{2}}=\frac{V_m}{R\sqrt{2}}\left(\sin ce,V_m=IR\right) \\ or{I}_{rms}=\frac{200}{10\times \sqrt{2}}=14.14 \\ or{I}_{rms}=14.14A. \end{gathered}$$

Therefore, the RMS current is $14.14A.$

Step 5: Finding the power dissipation

We know, power dissipation, $P_d=V_{rms}{I}_{rms}\cos \theta$. So,

$$\begin{gathered} P_d=V_{rms}{I}_{rms}\cos \theta =\left(141.42\right)\times \left(14.14\right)\times 1 \\ or{P}_d=1999.6watt. \end{gathered}$$

Therefore, the power dissipation as heat is $1999.6watt.$