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Question: For a transformer, the turn ratio \(3\) and its efficiency is \(0.75\). The current flowing in the p...

For a transformer, the turn ratio 33 and its efficiency is 0.750.75. The current flowing in the primary coil is 2A2 A and the voltage applied to it is 100V100 V. Then the voltage and current flowing in the secondary coil are __ respectively.
(A) 150V,1.5A150 V, 1.5 A
(B) 300V,0.5A300 V, 0.5 A
(C) 300V,1.5A300 V, 1.5 A
(D) 150V,0.5A150 V, 0.5 A

Explanation

Solution

Hint
The turn ratio is the ratio of either number of turns of secondary coil to primary coil or voltage of secondary coil to primary coil of the transformer. The efficiency of the transformer is the ratio of power of secondary coil to primary coil of the transformer.

Complete step by step answer
The turn ratio of transformer is r=3r=3, the efficiency of the transformer is η=0.75\eta = 0.75, the current in the primary coil of transformer is I1=2  A{I_1} = 2\;{\text{A}}and the voltage in the primary coil of transformer is V1=100  V{V_1} = 100\;{\text{V}}.
Use the expression of the turn ratio of the transformer to find the voltage of the secondary coil of the transformer. The expression is given as,
r=V2V1r = \dfrac{{{V_2}}}{{{V_1}}}
V2=rV1......(1){V_2} = r \cdot {V_1}......\left( 1 \right)
Substitute 33 for rr and 100  V100\;{\text{V}} for V1{V_1} in the equation (1) to find the voltage of the secondary transformer.
V2=(3)(100  V){V_2} = \left( 3 \right)\left( {100\;{\text{V}}} \right)
V2=300  V{V_2} = 300\;{\text{V}}
Use the expression of efficiency of the transformer to find the current in the secondary coil of the transformer. The expression is given as,
η=V2I2V1I1\eta = \dfrac{{{V_2} \cdot {I_2}}}{{{V_1} \cdot {I_1}}}
I2=ηV1I1V2  ......  (2){I_2} = \dfrac{{\eta \cdot {V_1} \cdot {I_1}}}{{{V_2}}}\;......\;\left( 2 \right)
Substitute 0.750.75 for η\eta , 100V100 V for V1{V_1}, 2A2 A for I1{I_1}and 300V300 V for V2{V_2}in the equation (2) to find the current flowing in the secondary coil of the transformer.
I2=(0.75)(100  V)(2  A)300  A{I_2} = \dfrac{{\left( {0.75} \right)\left( {100\;{\text{V}}} \right)\left( {2\;{\text{A}}} \right)}}{{300\;{\text{A}}}}
I2=0.5  A{I_2} = 0.5\;{\text{A}}
Therefore, the voltage of the secondary coil is 300V300 V and current flowing in the secondary coil is 0.5A0.5 A of the transformer.
Option (B) is correct.

Note
Be careful in using the expression of the turn ratio and efficiency because there is confusion in that the turn ratio is the ratio of the number of turns of primary to secondary or secondary to primary.