Question:
The number of turns in the primary and the secondary coil of an ideal transformer are 100 and 5000 respectively. If 3.3 kW power is supplied to the transformer at 220 V, find:
• (I) current in the primary coil
• (II) output voltage
The number of turns in the primary and the secondary coil of an ideal transformer are 100 and 5000 respectively. If 3.3 kW power is supplied to the transformer at 220 V, find:
• (I) current in the primary coil
• (II) output voltage
• (I) current in the primary coil
• (II) output voltage
Updated On: Jul 2, 2026
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Solution and Explanation
Given
\[ N_p=100,\qquad N_s=5000 \]
\[ V_p=220\;V \]
\[ P=3.3\;kW=3300\;W \]
Step 1: Calculate Current in the Primary Coil
For an ideal transformer,
\[ P=V_pI_p \]
Therefore,
\[ I_p=\frac{P}{V_p} \]
Substituting the given values,
\[ I_p=\frac{3300}{220} =15\;A \]
Hence,
\[ \boxed{I_p=15\;A} \]
Step 2: Calculate the Output Voltage
For an ideal transformer,
\[ \frac{V_s}{V_p}=\frac{N_s}{N_p} \]
Substituting the given values,
\[ \frac{V_s}{220}=\frac{5000}{100}=50 \]
Therefore,
\[ V_s=220\times50 =11000\;V \]
Hence,
\[ \boxed{V_s=11000\;V=11\;kV} \]
Final Answer
- (i) Current in the primary coil:
\[ \boxed{15\;A} \]
- (ii) Output voltage:
\[ \boxed{11000\;V=11\;kV} \]
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