An inductance of `2.0 H`, a capacitance of `18 muF` and a resistance of `10 kOmega` are connected to an AC source of `20 V` with adjustable frequency. (a) What frequency should be chosen to maximise the current in the circuit? (b) What is the value of this maximum current?

Answer

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Finding the Frequency to Maximize Current

1. **Identify the given values:**
- Inductance \( L = 2.0 \, \text{H} \)
- Capacitance \( C = 18 \, \mu\text{F} = 18 \times 10^{-6} \, \text{F} \)
- Resistance \( R = 10 \, \text{k}\Omega = 10 \times 10^{3} \, \Omega \)
- Voltage \( V = 20 \, \text{V} \)

2. **Determine the resonance condition:**
- At resonance, the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).
- The formula for resonance frequency \( f \) is given by:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]

3. **Substitute the values into the formula:**
\[ f = \frac{1}{2\pi \sqrt{2.0 \times 18 \times 10^{-6}}} \]

4. **Calculate the value inside the square root:**
\[ LC = 2.0 \times 18 \times 10^{-6} = 36 \times 10^{-6} = 3.6 \times 10^{-5} \]
\[ \sqrt{LC} = \sqrt{3.6 \times 10^{-5}} \approx 0.006 \]

5. **Calculate the frequency:**
\[ f = \frac{1}{2\pi \times 0.006} \approx \frac{1}{0.0377} \approx 26.5 \, \text{Hz} \]
Rounding off, we get:
\[ f \approx 27 \, \text{Hz} \]

### Part (b): Finding the Maximum Current

1. **Use Ohm's law to find the maximum current:**
- The maximum current \( I \) in the circuit can be calculated using:
\[ I = \frac{V}{R} \]

2. **Substitute the values:**
\[ I = \frac{20 \, \text{V}}{10 \times 10^{3} \, \Omega} = \frac{20}{10000} = 0.002 \, \text{A} \]

3. **Convert to milliampere:**
\[ I = 2 \, \text{mA} \]

### Final Answers:
- (a) The frequency to maximize the current is \( 27 \, \text{Hz} \).
- (b) The value of the maximum current is \( 2 \, \text{mA} \).