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, and (II) output voltage.

Hint:

Always rigorously check the units before starting any power calculation.
Converting kilowatts (kW) forcefully into standard watts (W) is a highly critical first step that students frequently overlook, causing massive decimal point errors.

Answer 1

Concept:
• An ideal transformer is assumed to be $100\%$ efficient, meaning absolutely zero power is lost during the transfer.
• Input electrical power perfectly equals output electrical power ($P_{in} = P_{out}$).
• Electrical power for AC circuits (assuming unity power factor for simple transformer problems) is $P = V \cdot I$.
• The voltage elegantly scales strictly according to the turns ratio equation: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.

Step 1: Identify the explicitly given parameters
Number of turns heavily wound in the primary coil: $N_p = 100$.
Number of turns heavily wound in the secondary coil: $N_s = 5000$.
The total input power rigorously supplied to the primary: $P = 3.3 \text{ kW} = 3300 \text{ W}$.
The applied primary voltage: $V_p = 220 \text{ V}$.

Step 2: Calculate the primary current (I)
The electrical power fundamentally delivered to the primary coil is completely determined by the product of primary voltage and primary current.
\[ P = V_p \cdot I_p \]
To find the primary current $I_p$, we simply rearrange the formula:
\[ I_p = \frac{P}{V_p} \]
Substitute the known values heavily into the arranged equation:
\[ I_p = \frac{3300 \text{ W}}{220 \text{ V}} \]
Simplify the fraction carefully by removing a zero:
\[ I_p = \frac{330}{22} \]
Divide thoroughly to get the exact value:
\[ I_p = 15 \text{ A} \]
The current powerfully drawn by the primary coil is therefore precisely $15 \text{ A}$.

Step 3: Calculate the output voltage (II)
To precisely find the output (secondary) voltage, we must employ the fundamental transformer turns ratio formula derived earlier.
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
We can dynamically rearrange this to solve specifically for the secondary voltage $V_s$:
\[ V_s = V_p \times \left( \frac{N_s}{N_p} \right) \]
Insert all the designated given values directly into this relation:
\[ V_s = 220 \times \left( \frac{5000}{100} \right) \]
First, neatly resolve the bracketed turns ratio, which signifies a massive step-up factor:
\[ \frac{5000}{100} = 50 \]
Now, multiply this substantial factor by the primary voltage:
\[ V_s = 220 \times 50 \]
\[ V_s = 11000 \text{ V} \]
This enormous output can also be written efficiently as $11 \text{ kV}$.
The final output voltage securely provided by the secondary coil is $11000 \text{ V}$.