NEET Electromagnetic Induction MCQs & PYQs

Practice NEET Electromagnetic Induction Questions

Question 141:

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 142:

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 143:

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 144:

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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