Question

A transformer has 600 turns in the primary coil and 9000 turns in the secondary coil if the transformer is 100% efficient and the current in the secondary coil is 0.15A, determine the current in the primary coil

Answer 1

Answer:

Sorry

Explanation:

Answer 2

The current in the primary coil is 2.25 A.

Explanation:

The efficiency of the transformer is given as

$\eta=\frac{P_o_u_t}{P_i_n} =\frac{V_sI_s}{V_pI_p}$

Here $V_s,V_p$ are the potential difference in the secondary and primary coil in volts respectively and $I_s,I_p$ are currents in the secondary and primary coils respectively.

As we know the ratio of potential differences on the transformer coils matches the ratio of the numbers of turns on the coils.

So, $V_s=9000V$  and $V_p=600 V$.

Given $I_s=0.15 A$ and  $\eta=100 \%$.

Substitute the values, we get

$100\%=\frac{9000\times0.15}{600\times I_p}$

$I_p=\frac{9000\times0.15}{600\times1}$

$I_p=2.25A$

Thus, the current in the primary coil is 2.25 A.

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