A transformer is used to light a 100 W, 110 V lamp from a 220 V supply. If the supply current is 0.6 A, the efficiency of the transformer is

Question

A transformer is used to light a 100 W, 110 V lamp from a 220 V supply. If the supply current is 0.6 A, the efficiency of the transformer is (A) 66% (B) 76% (C) 86% (D) 96%

Tardigrade Admin · Accepted Answer

As efficiency of the transformer, η =(Pout/Pin) Here. Pout = 100 W , : : Vin = 220 V Iin = 0.6 A ∴ : :: Pin = Vin × Iin = 220 × 0.6 = 132 W So, η = (100/132) = 0.7576 = 75.76 % = 76 %

Q. A transformer is used to light a $100 W$ , $110 V$ lamp from a $220 V$ supply. If the supply current is $0.6 A$ , the efficiency of the transformer is

66%

76%

86%

96%

As efficiency of the transformer,
$η = \frac{P _{o u t}}{P _{in}}$
Here. $P_{o u t} = 100 W, V_{in} = 220 V$
$I_{in} = 0.6 A$
$∴ P_{in} = V_{in} \times I_{in} = 220 \times 0.6 = 132 W$
So, $η = \frac{100}{132} = 0.7576 = 75.76% = 76%$

Q. A transformer is used to light a 100W, 110V lamp from a 220V supply. If the supply current is 0.6A, the efficiency of the transformer is

66%

76%

86%

96%

As efficiency of the transformer, η=Pin​Pout​​ Here. Pout​=100W,Vin​=220V Iin​=0.6A ∴Pin​=Vin​×Iin​=220×0.6=132W So, η=132100​=0.7576=75.76%=76%

Q. A transformer is used to light a $100 W$ , $110 V$ lamp from a $220 V$ supply. If the supply current is $0.6 A$ , the efficiency of the transformer is

As efficiency of the transformer,
$η = \frac{P _{o u t}}{P _{in}}$
Here. $P_{o u t} = 100 W, V_{in} = 220 V$
$I_{in} = 0.6 A$
$∴ P_{in} = V_{in} \times I_{in} = 220 \times 0.6 = 132 W$
So, $η = \frac{100}{132} = 0.7576 = 75.76% = 76%$