Question 340: AC fundamentals

Question

A transformer is used to set up an alternating e.m.f. of 220 V to 4.4 kV to transmit 6.6 kW of power. The primary coil has 1000 turns. The current rating of the secondary coil is (Transformer is ideal)

Accepted Answer

Number of turns of in primary and secondary coil be $\mathrm{N}_{\mathrm{p}}$ and $\mathrm{N}_{\mathrm{s}}$,
For ideal transformer,
$$\begin{aligned}
& \frac{N_s}{N_p}=\frac{V_s}{V_p} \
& N_s=\frac{4.4 	imes 10^3}{220} 	imes 1000=2 	imes 10^4
\end{aligned}$$
Power supply at primary coil is given as,
$$\begin{aligned}
& P=I_p V_p=6.6 	imes 10^3 \mathrm{~V} \
& I_p=\frac{6.6 	imes 10^3 V}{220}=30 \mathrm{~A}
\end{aligned}$$
For ideal transformer,
$$\begin{aligned}
& \frac{I_s}{I_p}=\frac{N_p}{N_s} \
& I_s=\frac{10^3}{2 	imes 10^4} 	imes 30=1.5 \mathrm{~A}
\end{aligned}$$

Answer

0.8 A

Answer

1.2 A

Answer

1.5 A

Answer

1.8 A

AC fundamentals

A transformer is used to set up an alternating e.m.f. of 220 V to 4.4 kV to transmit 6.6 kW of power. The primary coil has 1000 turns. The current rating of the secondary coil is (Transformer is ideal)

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