A transformer of efficiency 90% has turns ratio 1 : 10. If the voltage across the primary is 220 V and current in the primary is 0.5 A, then the current in secondary is

Options

A. 5.5 A
B. `5 A`
C. `4 A`
D. 4.5 A

Answer

To solve the problem step by step, we will use the information given about the transformer, including its efficiency, turns ratio, primary voltage, and primary current.

### Step 1: Understand the given data
- Efficiency (η) = 90% = 0.9
- Turns ratio (Np:Ns) = 1:10 (which means Np = 1 and Ns = 10)
- Voltage across primary (Vp) = 220 V
- Current in primary (Ip) = 0.5 A

### Step 2: Calculate the voltage across the secondary (Vs)
Using the turns ratio, we can express the relationship between primary and secondary voltages:
\[ \frac{Vs}{Vp} = \frac{Ns}{Np} \]
Substituting the known values:
\[ \frac{Vs}{220} = \frac{10}{1} \]
From this, we can find Vs:
\[ Vs = 220 \times 10 = 2200 \text{ V} \]

### Step 3: Calculate the power in the primary (Pp)
The power in the primary can be calculated using the formula:
\[ Pp = Vp \times Ip \]
Substituting the known values:
\[ Pp = 220 \times 0.5 = 110 \text{ W} \]

### Step 4: Calculate the power in the secondary (Ps) using efficiency
The efficiency of the transformer relates the power in the primary and secondary:
\[ \eta = \frac{Ps}{Pp} \]
Rearranging gives:
\[ Ps = \eta \times Pp \]
Substituting the values:
\[ Ps = 0.9 \times 110 = 99 \text{ W} \]

### Step 5: Calculate the current in the secondary (Is)
Using the power in the secondary, we can find the current in the secondary using the formula:
\[ Ps = Vs \times Is \]
Rearranging gives:
\[ Is = \frac{Ps}{Vs} \]
Substituting the known values:
\[ Is = \frac{99}{2200} = 0.045 \text{ A} \]

### Step 6: Final calculation of Is
Since we need to find the current in the secondary, we can also use the relationship between primary and secondary currents given by the turns ratio:
\[ \frac{Is}{Ip} = \frac{Np}{Ns} \]
Substituting the known values:
\[ \frac{Is}{0.5} = \frac{1}{10} \]
From this, we find:
\[ Is = 0.5 \times \frac{1}{10} = 0.05 \text{ A} \]

### Conclusion
The current in the secondary (Is) is 0.05 A.