Across a 100 V, 50 Hz supply, a coil with a resistance of 10 and an inductance of 210 mH is connected in series with a variable capacitor. Calculate:

Given:
Capacitance $C = 58 \, \mu\text{F} = 58 \times 10^{-6} \, \text{F}$
Resistance $R = 15 \, \Omega$
Inductance $L = 128 \, \text{mH} = 128 \times 10^{-3} \, \text{H}$
Supply voltage $V = 100 \, \text{V}$ (RMS, assumed)

(Since I cannot draw an image, I will describe the key features of the phasor diagram at resonance).

• The current phasor ($I$) is typically drawn along the positive x-axis (reference).
• The voltage across the resistor ($V_R = IR$) is in phase with the current, so $V_R$ is also along the positive x-axis.
• The voltage across the inductor ($V_L = IX_L$) leads the current by $90^\circ$, so $V_L$ points along the positive y-axis.
• The voltage across the capacitor ($V_C = IX_C$) lags the current by $90^\circ$, so $V_C$ points along the negative y-axis.
• At resonance, $V_L$ and $V_C$ are equal in magnitude and opposite in direction, so they cancel each other out.
• The total supply voltage ($V$) is equal to $V_R$ and is in phase with the current. The diagram would show $I$ and $V_R$ on the positive x-axis, $V_L$ on the positive y-axis, and $V_C$ on the negative y-axis, with $V_L$ and $V_C$ having equal lengths. The resultant voltage $V$ would coincide with $V_R$.

Given:
Coil resistance $R = 15 \, \Omega$
Coil inductance $L = 200 \, \text{mH} = 200 \times 10^{-3} \, \text{H}$

Assumption: This question implies a parallel resonant circuit where the coil (R and L in series) is connected in parallel with a capacitor. However, the value of the capacitor is not provided in the question. Without the capacitor value, it is not possible to calculate a specific non-zero resonant frequency.
If the circuit consists only of the coil (R and L in series) connected to the source, the current will always lag the voltage at any non-zero frequency. The current would only be in phase with the voltage if the circuit were purely resistive, which means the inductance $L$ would have to be zero or the frequency would have to be zero (DC).

Therefore, based on the information given, this question cannot be solved for a non-zero frequency.
Missing information: Value of the parallel capacitor.

The Q factor (Quality factor) for a parallel resonant circuit where the coil (R and L in series) is in parallel with a capacitor is given by:

or

Since the resonant frequency $\omega_0$ (or $f_0$) cannot be determined without the capacitor value (as explained in 2.1), the Q factor also cannot be calculated.
Missing information: Value of the parallel capacitor (and thus resonant frequency).

Given:
Supply voltage $V = 100 \, \text{V}$ (RMS, assumed)
Supply frequency $f = 50 \, \text{Hz}$
Coil resistance $R = 10 \, \Omega$
Coil inductance $L = 210 \, \text{mH} = 210 \times 10^{-3} \, \text{H}$
A variable capacitor is connected in series.

This is the same as Question 3.1.
The value of the capacitor is $\boxed{48.24 \, \mu\text{F}}$.