A pure inductance of 1.0H is connected across a 110V, 70Hz Source. Find the (a) reactanace, (b) current and (c ) peak value of current.

Options

A. reactance of the circuit is `440 Omega`
B. current of the circuit is `0.25 Omega`
C. reactance of the circuit is `880 Omega`
D. current of the circuit is `0.5 Omega`

Answer

To solve the problem step by step, we will find the reactance, current, and peak value of current for a pure inductance connected across an AC source.

### Given:
- Inductance (L) = 1.0 H
- RMS Voltage (V_rms) = 110 V
- Frequency (f) = 70 Hz

### Step 1: Calculate the Reactance (X_L)
The reactance of an inductor is given by the formula:
\[ X_L = \omega L \]
where \( \omega = 2 \pi f \).

First, we calculate \( \omega \):
\[ \omega = 2 \pi f = 2 \pi \times 70 \]

Calculating \( \omega \):
\[ \omega \approx 2 \times 3.14 \times 70 \approx 439.82 \, \text{rad/s} \]

Now, substitute \( \omega \) and \( L \) into the reactance formula:
\[ X_L = 439.82 \times 1.0 \approx 439.82 \, \Omega \]

### Step 2: Calculate the Current (I_rms)
The RMS current can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{X_L} \]

Substituting the values:
\[ I_{rms} = \frac{110}{439.82} \approx 0.25 \, \text{A} \]

### Step 3: Calculate the Peak Value of Current (I_0)
The peak current can be calculated from the RMS current using the relationship:
\[ I_0 = \sqrt{2} \times I_{rms} \]

Substituting the value of \( I_{rms} \):
\[ I_0 = \sqrt{2} \times 0.25 \]

Calculating \( I_0 \):
\[ I_0 \approx 0.353 \, \text{A} \]

### Final Answers:
(a) Reactance \( X_L \approx 439.82 \, \Omega \)
(b) RMS Current \( I_{rms} \approx 0.25 \, \text{A} \)
(c) Peak Current \( I_0 \approx 0.353 \, \text{A} \)