Question 346: AC fundamentals

Question

A transformer having efficiency $90 \%$ is working on 200 V and 3 kw power supply. If the current in the secondary coil is 6 A , the voltage across the secondary coil and the current in the primary coil are respectively

Accepted Answer

The efficiency of a transformer is given by the formula:
$$ 	ext{Efficiency} (\eta) = \frac{	ext{Output Power}}{	ext{Input Power}} 	imes 100\% $$
Given the efficiency $\eta = 90\%$, the input power $P_{	ext{in}} = 3 \, 	ext{kW} = 3000 \, 	ext{W}$, and the primary voltage $V_p = 200 \, 	ext{V}$, we can first find the output power $P_{	ext{out}}$:
$$ P_{	ext{out}} = \eta 	imes P_{	ext{in}} = 0.9 	imes 3000 \, 	ext{W} = 2700 \, 	ext{W} $$
The current in the secondary coil is given as $I_s = 6 \, 	ext{A}$. The voltage across the secondary coil $V_s$ can be found using the formula:
$$ P_{	ext{out}} = V_s 	imes I_s $$
Substitute the known values:
$$ 2700 \, 	ext{W} = V_s 	imes 6 \, 	ext{A} $$
Solving for $V_s$:
$$ V_s = \frac{2700 \, 	ext{W}}{6 \, 	ext{A}} = 450 \, 	ext{V} $$
Next, to find the current in the primary coil $I_p$, use the formula related to the input power $P_{	ext{in}}$:
$$ P_{	ext{in}} = V_p 	imes I_p $$
Substitute the known values:
$$ 3000 \, 	ext{W} = 200 \, 	ext{V} 	imes I_p $$
Solving for $I_p$:
$$ I_p = \frac{3000 \, 	ext{W}}{200 \, 	ext{V}} = 15 \, 	ext{A} $$
Thus, the voltage across the secondary coil and the current in the primary coil are $450 \, 	ext{V}$ and $15 \, 	ext{A}$, respectively. The correct option is:
Option B
: $450 \, 	ext{V}, 15 \, 	ext{A}$

Answer

$300 \mathrm{~V}, 15 \mathrm{~A}$

Answer

$450 \mathrm{~V}, 15 \mathrm{~A}$

Answer

$450 \mathrm{~V}, 13 \cdot 5 \mathrm{~A}$

Answer

$600 \mathrm{~V}, 15 \mathrm{~A}$

AC fundamentals

A transformer having efficiency 90 \% is working on 200 V and 3 kw power supply. If the current in the secondary coil is 6 A , the voltage across the secondary coil and the current in the primary coil are respectively

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