MHT CET · Physics · Alternating Current
A transformer has 120 turns in the primary coil and carries 5 A current. Input power is one kilowatt. To have 560 V output, the number of turns in secondary coil will be
- A 168
- B 200
- C 336
- D 400
Answer & Solution
Correct Answer
(C) 336
Step-by-step Solution
Detailed explanation
\(\therefore \frac{N_s}{N_p}=\frac{V_s}{V_p} \)
\( P=V I \)
\( \left.1000=V_p \times 5 \text {...(given } P=1 \mathrm{~kW}, \mathrm{I}=5 \mathrm{~A}\right) \)
\( \mathrm{V}_{\mathrm{p}}=\frac{1000}{5}=200 \mathrm{~V} \)
\( \therefore \frac{\mathrm{~N}_{\mathrm{s}}}{120}=\frac{560}{200} \)
\( \therefore \mathrm{~N}_{\mathrm{s}}=\frac{120 \times 560}{200}=336\)
\( P=V I \)
\( \left.1000=V_p \times 5 \text {...(given } P=1 \mathrm{~kW}, \mathrm{I}=5 \mathrm{~A}\right) \)
\( \mathrm{V}_{\mathrm{p}}=\frac{1000}{5}=200 \mathrm{~V} \)
\( \therefore \frac{\mathrm{~N}_{\mathrm{s}}}{120}=\frac{560}{200} \)
\( \therefore \mathrm{~N}_{\mathrm{s}}=\frac{120 \times 560}{200}=336\)
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