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Question: What will be the secondary current in the certain loaded transformer if the secondary voltage is one...

What will be the secondary current in the certain loaded transformer if the secondary voltage is one – fourth the primary voltage?
A) One – fourth the primary current
B) four times the primary current
C) equal to the primary current
D) one – fourth the primary current and equal to the primary current

Explanation

Solution

Use the formula of turn’s ratio which gives us the relation between number of windings, voltage and current in transformer for primary coil and secondary coil respectively. Also use the relation between primary voltage and secondary voltage as given in the question.

Complete step by step answer:
The electrical devices which consist of two or more than two coils of wire. It is used to transfer electrical energy by means of changing magnetic fields. Input is given through the primary coil of a transformer and output is taken from the secondary coil of a transformer. Generally, the core of a transformer is formed by the insulator.
On the basis of number of turns, the transformer can be classified into two –

  1. Step Up, and
  2. Step Down
    If the secondary coil of the transformer has more turns than the primary coil then, the transformer is said to be Step – Up transformer. The induced EMF is greater than the input signal.
    If the primary winding of the transformer has more number of turns than secondary winding of transformer then, this transformer is said to be Step – Down transformer. So, the input signal becomes greater than the induced EMF.
    Now, according to the question, it is given that secondary voltage is one – fourth the primary voltage.
    Therefore, let pp be the primary winding and ss be the secondary winding of the transformer.
    So,
    Vs=Vp4 VpVs=4(1)  {V_s} = \dfrac{{{V_p}}}{4} \\\ \dfrac{{{V_p}}}{{{V_s}}} = 4 \cdots \left( 1 \right) \\\
    We know that, turns ratio can be expressed as –
    NpNs=VpVs=IsIp\dfrac{{{N_p}}}{{{N_s}}} = \dfrac{{{V_p}}}{{{V_s}}} = \dfrac{{{I_s}}}{{{I_p}}}
    Using equation (1)\left( 1 \right), we get –
    4=IsIp Is=4Ip  4 = \dfrac{{{I_s}}}{{{I_p}}} \\\ {I_s} = 4{I_p} \\\
    Hence, secondary current becomes four times the primary current.
    Therefore, option (B) is the correct answer.

Note: As the number of turns of primary and secondary windings affect the voltage ratings, it is important to maintain a ratio between the turns so as to have an idea regarding the voltages induced.