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.