Question

A coil with a resistance of 6,75ohm abd an Inductance of 60mH is connected in series to a variable capacitor. This circuit is then connected across a 120, 50Hz supply.
1) calculate the value of the capacitor that must be adjusted to give resonance
2)calculate the voltage drop across the coil

Answer 1

To solve the given problem, we need to calculate two things.

1. The value of the capacitor that needs to be adjusted for resonance.
2. The voltage drop across the coil.

### Step 1: Resonance Condition

In a series RLC circuit, resonance occurs when the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\).

1. **Calculate the Inductive Reactance (\(X_L\))**:
The inductive reactance is given by:
\[
X_L = 2 \pi f L
\]
where:
- \(f\) is the frequency (50 Hz),
- \(L\) is the inductance (60 mH or \(60 \times 10^{-3} \, H\)).

\[
X_L = 2 \pi (50) (60 \times 10^{-3}) = 2 \pi (50) (0.06) \approx 18.85 \, \Omega
\]

2. **Set \(X_L\) equal to \(X_C\)** for resonance:
The capacitive reactance \(X_C\) is given by:
\[
X_C = \frac{1}{2 \pi f C}
\]

Setting these equal gives:
\[
18.85 = \frac{1}{2 \pi (50) C}
\]

Rearranging for \(C\):
\[
C = \frac{1}{2 \pi (50)(18.85)}
\]

Now calculate \(C\):
\[
C = \frac{1}{2 \pi (50)(18.85)} = \frac{1}{3141.59} \approx 0.0003183 \, F
\]
or
\[
C \approx 318.3 \, \mu F
\]

### Step 2: Voltage Drop Across the Coil

The voltage drop across the coil (which has resistance \(R\)) can be calculated using Ohm's Law:
\[
V_R = I \cdot R
\]

First, we need to find the current \(I\) in the circuit. At resonance, the total impedance \(Z\) of the circuit is simply the resistance \(R\) since \(X_L\) and \(X_C\) cancel each other out.

The total voltage across the circuit is given as \(120 V\) (RMS). The current \(I\) is calculated as:
\[
I = \frac{V}{R}
\]

Using:
- \(V = 120 \, V\)
- \(R = 6.75 \, \Omega\)

Calculating \(I\):
\[
I = \frac{120}{6.75} \approx 17.78 \, A
\]

Now, substitute this current into the expression for the voltage drop across the coil:
\[
V_R = I \cdot R = 17.78 \cdot 6.75 \approx 119.4 \, V
\]

### Summary of Results

1. The adjusted capacitor value for resonance is approximately \(318.3 \, \mu F\).
2. The voltage drop across the coil is approximately \(119.4 \, V\).