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Alternating Current question

2002 · Shift 0 · Q128
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Alternating Current question

2002 · Shift 0 · Q128

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
In a transformer, number of turns in the primary coil are 140140140 and that in the secondary coil are 280.280.280. If current in primary coil is 4A,4A,4A, then that in the secondary coil is
  1. A
    4A4A4A
  2. B
    2A2A2A
  3. C
    6A6A6A
  4. D
    10A10A10A
View written solutionFree

Correct answer: B

  1. Use the ideal transformer relations

    For an ideal transformer, VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}Vp​Vs​​=Np​Ns​​ and since input power equals output power, VpIp=VsIsV_p I_p = V_s I_sVp​Ip​=Vs​Is​

  2. Relate current to number of turns

    From the above relations, IsIp=NpNs\frac{I_s}{I_p} = \frac{N_p}{N_s}Ip​Is​​=Ns​Np​​

  3. Substitute the given values

    Given: Np=140,Ns=280,Ip=4 AN_p = 140, \quad N_s = 280, \quad I_p = 4\text{ A}Np​=140,Ns​=280,Ip​=4 A

    So, Is=Ip⋅NpNs=4⋅140280I_s = I_p \cdot \frac{N_p}{N_s} = 4 \cdot \frac{140}{280}Is​=Ip​⋅Ns​Np​​=4⋅280140​

  4. Calculate

    Is=4⋅12=2 AI_s = 4 \cdot \frac{1}{2} = 2\text{ A}Is​=4⋅21​=2 A

  5. Check options

    • A: 4 A4\text{ A}4 A ❌
    • B: 2 A2\text{ A}2 A ✅
    • C: 6 A6\text{ A}6 A ❌
    • D: 10 A10\text{ A}10 A ❌

Therefore, the current in the secondary coil is 2 A\boxed{2\text{ A}}2 A​

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