A coil of inductance 0.12 H and resistance 4\(\Omega\), is connected across a 240V, 50Hz supply. Calculate the current through it. [\(\pi\) = 3.142]

Answer Details

We can use the formula for the current in an AC circuit with an inductor and a resistor: \begin{equation*} I = \frac{V}{\sqrt{R^2 + (\omega L)^2}} \end{equation*} where V is the voltage, R is the resistance, L is the inductance, and \(\omega\) is the angular frequency. We can find \(\omega\) using the formula: \begin{equation*} \omega = 2\pi f \end{equation*} where f is the frequency. Substituting the given values, we get: \begin{align*} \omega &= 2\pi \times 50\text{ Hz} \\ &= 100\pi\text{ rad/s} \end{align*} Now we can substitute all the values into the first formula: \begin{align*} I &= \frac{V}{\sqrt{R^2 + (\omega L)^2}} \\ &= \frac{240\text{ V}}{\sqrt{(4\Omega)^2 + (100\pi \text{ rad/s} \times 0.12\text{ H})^2}} \\ &\approx 6.3\text{ A} \end{align*} Therefore, the current through the coil is approximately 6.3A. Answer is correct.