Question

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

(1) 1.59 A 

(2) 13.33 A 

(3) 1.33 A 

(4) 15.1 A

Answer 1

(2) 13.33 A 

Efficiency \(\eta\) = 80%, P_i = 4kW = 4000W

\(V_p = 10V , V_s = 240V\)

For primary coil : \(I_p = \frac {P_i}{V_p} = \frac {4000}{10} = 400A\)

\(\eta = \frac {V_s I _s}{V_p I _p} \Rightarrow \frac {80}{100} = \frac {240I_s}{10 \times 400}\)

\(\Rightarrow I_s = 13.33 A\)