A transformer with efficiency 80 \% works at 4 ~kW and 100 ~V . If the secondary voltage is 200 ~V , then the primary and secondary currents are respectively

Correct answer: A. 40 ~A and 16 ~A aligned & (a) : Here, =80 \% & P_i=4 kW=4000 W & V_p=100 V, V_s=200 V & I_p=(P_i)/(V_p)=(4000)/(100)=40 A & =(V_s I_s)/(V_p I_p) ; (80)/(100)=(200 I_s)/(4000)=(I_s)/(20) & or I_s=(20 × 80)/(100)=16 A aligned

AIIMS 2017

Exam: AIIMS
Year: 2017
Subject: Physics
Chapter: Alternating Current

AIIMS 2017 · Physics · Alternating Current

A transformer with efficiency 80 \% works at 4 ~kW and 100 ~V . If...

A transformer with efficiency

80 \%

works at

4 \mathrm{~kW}

and

100 \mathrm{~V}

. If the secondary voltage is

200 \mathrm{~V}

, then the primary and secondary currents are respectively

Options

A $40 \mathrm{~A}$ and $16 \mathrm{~A}$
B $16 \mathrm{~A}$ and $40 \mathrm{~A}$
C $20 \mathrm{~A}$ and $40 \mathrm{~A}$
D $40 \mathrm{~A}$ and $20 \mathrm{~A}$

Correct answer

A. $40 \mathrm{~A}$ and $16 \mathrm{~A}$

Step-by-step solution

\begin{aligned} & \text{ (a) : Here, } \eta=80 \% \\ & P_i=4 \mathrm{ kW}=4000 \mathrm{ W} \\ & V_p=100 \mathrm{ V}, V_s=200 \mathrm{ V} \\ & I_p=\frac{P_i}{V_p}=\frac{4000}{100}=40 \mathrm{ A} \\ & \eta=\frac{V_s I_s}{V_p I_p} ; \frac{80}{100}=\frac{200 I_s}{4000}=\frac{I_s}{20} \\ & \text{ or } I_s=\frac{20 \times 80}{100}=16 \mathrm{ A}\end{aligned}

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