A transformer having efficiency of 75 % is working on 220 V and 4.4 kW power supply. If the current in the secondary coil is 5 A. What will be the voltage across secondary coil and the current in primary coil?

V_s = 220 V, i_p = 20A

V_s = 660 V, i_p = 15 A

V_s = 660 V, i_p = 20 A

V_s = 220 V, i_p = 15 A

Solution

V_s = 660 V, i_p = 20 A

Given:- P_in = 4.4 kW,

V_p = 220 V,

i_s = 5A,

V_s = ?

We know that

i_P = $\frac{P_{in}}{V_{P}}$

= $\frac{4.4 \times 10^{3}}{220}$

i_p = 20A

∴ Efficiency, $η = \frac{output power}{input power}$

Percentage efficiency

η = $\frac{P_{out}}{P_{in}} \times 100$

75% = 100 × $\frac{P_{out}}{4.4 \times 10^{3}}$

P_out = $\frac{75 \times 4.4 \times 10^{3}}{100}$

= 3300 W

∴ P_out = V_S. i_S

⇒ V_S = $\frac{3300}{5}$

⇒ V_S = 660 V

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