Question 3.35: A resistor, a variable iron-core inductor, and a capacitor a......
A resistor, a variable iron-core inductor, and a capacitor are connected across a 230 V, 50 Hz supply. By varying the position of the iron core inside the inductor coil, its inductance is changed. Maximum current of 1.5 A was obtained in the circuit by changing the inductance of the coil. At that time the voltage across the capacitor was measured as 600 V. Calculate the values of circuit parameters.
We know that the maximum current flows at resonance when X_L = X_C and Z = R
Maximum current, I_0=I_m=\frac{V}{Z}=\frac{V}{R}[\because Z=R]
Therefore, \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{m}}}=\frac{230}{1.5}=153.3 \Omega.
\begin{aligned}\text{At resonance, }\quad \mathrm{V}_{\mathrm{L}} & =\mathrm{V}_{\mathrm{C}}=600 \mathrm{~V} \\ \mathrm{~V}_{\mathrm{L}} & =\mathrm{I}_{\mathrm{m}} \mathrm{X}_{\mathrm{L}}=600 \mathrm{~V} \\ \mathrm{X}_{\mathrm{L}} & =\frac{600}{1.5}=400 \Omega \\\text{or, }\quad 2 \pi \mathrm{fL} & =400 \\ \mathrm{~L} & =\frac{400}{6.28 \times 50}=1.27 \mathrm{H} \\ \mathrm{X}_{\mathrm{C}} & =\mathrm{X}_{\mathrm{L}}=400 \end{aligned}or, \frac{1}{2 \pi \mathrm{fC}}=400
or, \mathrm{C}=\frac{1}{314 \times 400}=7.96 \times 10^{-6} \mathrm{~F}