Question
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
40 A and 16 A
16 A and 40 A
20 A and 40 A
40 A and 20 A
Solution
A.
40 A and 16 A
Given:-
η = 80%
Pi = 4 kW
⇒ Pi = 4000 W
Vp = 100 V
Vs = 200 V
Ip =
Ip = 40 A
The efficiency of the transformer is defined as the ratio of useful power output the input power, the two being measured in the same unit
Transformer efficiency given by,
η =
Where Vs → Secondary voltage
Vp → Primary voltage
Is → secondary current
Ip → primary current
⇒
=
⇒ Is =
⇒ Is = 16 A
