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A transformer is used to reduce the mains supply of 220 V to 11 V. If the currents in primary and secondary coils are 5 A and 90 A respectively, efficiency of transformer is
  • a)
    90%
  • b)
    75%
  • c)
    40%
  • d)
    60%
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A transformer is used to reduce the mains supply of 220 V to 11 V. If ...
HT side Power = Vp(Primary Voltage) x Ip(Primary Current) = 220 x 5 = 1100 W

LT side Power = Vs(Secondary Voltage) x Is(Secondary current) = 11 x 90 = 990 W

Loss = HT side Power -  LT side Power = 1100-990 = 110 W Or Loss(in %) = (Loss/HT side Power) = 110/1100 = 10%

Efficiency = 100-10 = 90%
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A transformer is used to reduce the mains supply of 220 V to 11 V. If ...
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A transformer is used to reduce the mains supply of 220 V to 11 V. If ...
Given:
Primary voltage (Vp) = 220 V
Secondary voltage (Vs) = 11 V
Primary current (Ip) = 5 A
Secondary current (Is) = 90 A

To find:
Efficiency of the transformer

Explanation:
Efficiency of a transformer is given by the formula:
Efficiency = (Output power / Input power) × 100%

The power in the primary coil (Input power) is given by the formula:
Input power = Vp × Ip

Similarly, the power in the secondary coil (Output power) is given by the formula:
Output power = Vs × Is

Let's calculate the input power and output power:

1. Input power:
Input power = Vp × Ip
Input power = 220 V × 5 A
Input power = 1100 W

2. Output power:
Output power = Vs × Is
Output power = 11 V × 90 A
Output power = 990 W

Now, let's calculate the efficiency of the transformer:

Efficiency = (Output power / Input power) × 100%
Efficiency = (990 W / 1100 W) × 100%
Efficiency ≈ 0.9 × 100%
Efficiency ≈ 90%

Therefore, the efficiency of the transformer is 90%.
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A transformer is used to reduce the mains supply of 220 V to 11 V. If the currents in primary and secondary coils are 5 A and 90 A respectively, efficiency of transformer isa)90%b)75%c)40%d)60%Correct answer is option 'A'. Can you explain this answer?
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