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An electric lamp having resistance 5 Ome...

An electric lamp having resistance 5 `Omega` gives correct brightness when 10 A current flows through it. We need to operate this lamp using AC source rated as 200 V-50 Hz. Calculate the self-indcutance of choke coil needed.

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To solve the problem step by step, we need to find the self-inductance of the choke coil required to operate the lamp at 200 V AC while ensuring that the current remains at 10 A. ### Step 1: Calculate the required impedance (Z) The impedance of the circuit can be calculated using Ohm's law, which states that: \[ Z = \frac{V}{I} \] Where: - \( V = 200 \, \text{V} \) (the voltage of the AC source) - \( I = 10 \, \text{A} \) (the desired current) ...
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Knowledge Check

  • When 100 V dc is applied a coil, a current of 1 A flows through it. When 100 V 50 Hz ac is applied to the same coil, only 0.5 A flows. The indcutance of the coil is

    A
    5.5 mH
    B
    0.55 mH
    C
    55 mH
    D
    0.55 H
  • A moving coil galvanometer of resistance 50 Omega gives a full scale deflection , when a current of 0.5 m A is passed through it. To convert it to a voltmeter of range , 10 V , the resistance required to be placed in series is

    A
    `1995 Omega`
    B
    `2000 Omega`
    C
    `10000 Omega`
    D
    `19950 Omega`
  • A choke coil having a resistance of 10Omega and inductance of 0.5 H is connected in series with a capacitor of 2.5 muF . The whole circuit has been connected to 230 V, 50 Hz supply. Calculate the value of current flowing through the circuit.

    A
    0.487 A
    B
    0.308 A
    C
    0.206 A
    D
    None of these
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