NEET Electromagnetic Induction MCQs & PYQs

Practice NEET Electromagnetic Induction Questions

Question 101:

easy

A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is:

(2010 Pre)

Efficiency of a transformer is given by $\eta = \frac{V_s I_s}{V_p I_p}$.
Substituting the given values: $0.80 = \frac{440 \times 2.0}{220 \times I_p}$.
Rearranging for primary current gives $I_p = \frac{880}{220 \times 0.8} = \frac{4}{0.8} = 5.0 A$.

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Question 102:

easy

A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 amp, the efficiency of the transformer is approximately:

(2007)

Output power is $P_{out} = 100 W$.
Input power is $P_{in} = V_p I_p = 220 \times 0.5 = 110 W$.
Efficiency $\eta = \frac{P_{out}}{P_{in}} = \frac{100}{110} = 0.909$, which is approximately $90\%$.

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Question 103:

easy

The primary and secondary coil of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi = \phi_0 – 4t$, where $\phi$ is in Webers, t is time in seconds and $\phi_0$ is a constant, the output voltage across the secondary coil is:

(2007)

Voltage across primary is $V_p = \left| \frac{d\phi}{dt} \right| = \left| \frac{d}{dt}(\phi_0 - 4t) \right| = 4 V$.
Using the transformer ratio $V_s = V_p \left(\frac{N_s}{N_p}\right)$.
Secondary voltage $V_s = 4 \times \left(\frac{1500}{50}\right) = 4 \times 30 = 120 V$.

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Question 104:

easy

The core of a transformer is laminated because:

(2006)

A solid metal core allows large eddy currents, which lead to significant energy loss as heat.
By laminating the core, its electrical resistance is increased.
This restricts the paths of the eddy currents and minimizes the associated energy losses.

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Question 105:

easy

Turn ratio of a step-up transformer is 1 : 25. If current in load coil is 2 A, then the current in primary coil will be:

(1998)

For an ideal transformer, the current ratio is inversely proportional to the turns ratio: $\frac{I_p}{I_s} = \frac{N_s}{N_p}$.
The step-up turns ratio is $N_p : N_s = 1 : 25$, so $\frac{N_s}{N_p} = 25$.
Primary current $I_p = I_s \times 25 = 2 \times 25 = 50 A$.

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Question 106:

easy

The primary winding of a transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an A.C. supply 20 V, 50 Hz. The secondary will have an output of

(1997)

Output voltage $V_s = V_p \left(\frac{N_s}{N_p}\right) = 20 \times \left(\frac{5000}{500}\right) = 200 V$.
The frequency of an alternating current does not change when passing through a transformer.
Therefore, the output will be $200 V, 50 Hz$.

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