In an A.C. circuit, the current flowing is \(I=5sin(100 t - \frac{π}{2}) ampere\) and the potential difference is \(V = 200 sin (100 t) volts\). The power consumption is equal to

Question

In an a.c. circuit, voltage and current are given by: \[ V = 100 \sin(100t) \, \text{V} \quad \text{and} \quad I = 100 \sin\left(100t + \frac{\pi}{3}\right) \, \text{mA}. \] The average power dissipated in one cycle is:

Answer 1

To determine the power consumption in the given A.C. circuit, we must first understand the phase relationship between the current and the voltage.

The expressions for current \((I)\) and voltage \((V)\) are given as:

Current: \(I = 5 \sin(100t - \frac{\pi}{2})\) amperes
Voltage: \(V = 200 \sin (100t)\) volts

The phase difference between the current and voltage can be found by comparing their expressions. The current has a phase of \(-\frac{\pi}{2}\) compared to the voltage, which has a phase of 0. This means our current lags the voltage by \(\frac{\pi}{2}\) radians (or 90 degrees).

The power consumed in an A.C. circuit is determined by the formula for real power:

\(P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi\)

Where:

\(V_{\text{rms}}\) and \(I_{\text{rms}}\) are the root mean square values of voltage and current respectively.
\(\phi\) is the phase difference between voltage and current.

Since the phase difference \(\phi = \frac{\pi}{2}\), we have:

\(\cos \frac{\pi}{2} = 0\)

As a result, the real power is:

\(P = V_{\text{rms}} \times I_{\text{rms}} \times 0 = 0 \text{ W}\)

Thus, the power consumption in this A.C. circuit is 0 W, which corresponds to the given correct answer.

Answer 2