Question 67: Electromagnetic Induction

Question

A transformer has an efficiency of $80 \%$ and works at $10 \mathrm{~V}$ and $4 \mathrm{~kW}$. If the secondary voltage is $240 \mathrm{~V}$, then the current in the secondary coil is :

Accepted Answer

To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency. The efficiency ($$ \eta $$) of the transformer is given by the ratio of the output power ($$ P_{	ext{out}} $$) to the input power ($$ P_{	ext{in}} $$) times 100%.
The given efficiency is $$ \eta = 80\% $$ or $$ \eta = 0.8 $$ in decimal form. The input power is also given as $$ P_{	ext{in}} = 4 	ext{kW} $$ or $$ P_{	ext{in}} = 4000 	ext{W} $$.
Let's calculate $$ P_{	ext{out}} $$ using the efficiency formula:
$$
P_{	ext{out}} = \eta 	imes P_{	ext{in}} = 0.8 	imes 4000 	ext{W} = 3200 	ext{W}
$$
Now that we have $$ P_{	ext{out}} $$, we can calculate the secondary current ($$ I_{	ext{secondary}} $$) using the formula:
$$
P_{	ext{out}} = V_{	ext{secondary}} 	imes I_{	ext{secondary}}
$$
We are given $$ V_{	ext{secondary}} = 240 	ext{V} $$.
Isolating $$ I_{	ext{secondary}} $$ gives us:
$$
I_{	ext{secondary}} = \frac{P_{	ext{out}}}{V_{	ext{secondary}}}
$$
Substituting the known values:
$$
I_{	ext{secondary}} = \frac{3200 	ext{W}}{240 	ext{V}}
$$
$$
I_{	ext{secondary}} = \frac{3200}{240}
$$
$$
I_{	ext{secondary}} = 13.33 	ext{A}
$$
Therefore, the current in the secondary coil is $$ 13.33 	ext{A} $$, which corresponds to Option B.

Answer

$1.33 \mathrm{~A}$

Answer

$13.33 \mathrm{~A}$

Answer

$1.59 \mathrm{~A}$

Answer

$15.1 \mathrm{~A}$

Electromagnetic Induction

A transformer has an efficiency of 80 \% and works at 10 \mathrm{~V} and 4 \mathrm{~kW}. If the secondary voltage is 240 \mathrm{~V}, then the current in the secondary coil is :

Choose the correct answer: