A transformer steps down from 220V to 22V with secondary impedance of `220Omega`. Current is drawn by a primary coil is

Options

A. `10^(-6)A`
B. `10^(-4)A`
C. `10^(-2)A`
D. `10^(2)A`

Explanation

To solve the problem, we need to find the current drawn by the primary coil (Ip) of a transformer that steps down voltage from 220V to 22V, given that the secondary impedance (Rs) is 220Ω.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Primary Voltage (Vp) = 220V
- Secondary Voltage (Vs) = 22V
- Secondary Impedance (Rs) = 220Ω

2. **Calculate the Secondary Current (Is):**
- We can find the secondary current using Ohm's Law, which states that \( I = \frac{V}{R} \).
- Here, \( Is = \frac{Vs}{Rs} \).
- Substitute the values:
\[ Is = \frac{22V}{220Ω} = \frac{1}{10} A = 0.1 A \]

3. **Use the Transformer Equation:**
- For an ideal transformer, the relationship between primary and secondary currents is given by:
\[ \frac{Vs}{Vp} = \frac{Ip}{Is} \]
- Rearranging this equation to find the primary current (Ip):
\[ Ip = \frac{Vs}{Vp} \times Is \]

4. **Substitute the Values:**
- Substitute \( Vs = 22V \), \( Vp = 220V \), and \( Is = 0.1A \):
\[ Ip = \frac{22V}{220V} \times 0.1A \]
- Simplifying this:
\[ Ip = \frac{1}{10} \times 0.1A = \frac{0.1}{10} A = 0.01 A \]

5. **Final Result:**
- Therefore, the current drawn by the primary coil (Ip) is:
\[ Ip = 0.01 A = 10^{-2} A \]

### Final Answer:
The current drawn by the primary coil is \( 0.01 A \) or \( 10^{-2} A \).

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