JEE Main Alternating Current PYQ

6. (JEE Main 2022 (Online) 25th June Morning Shift)

Match List-I with List-II.

	List - I		List -II
(A)	AC generator	(I)	Detects the presence of current in the circuit
(B)	Galvanometer	(II)	Converts mechanical energy into electrical energy
(C)	Transformer	(III)	Works on the principle of resonance in AC circuit
(D)	Metal detector	(IV)	Changes an alternating voltage for smaller or greater value

Choose the correct answer from the options given below

A. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

B. (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

C. (A) - (III), (B) - (IV), (C) - (II), (D) - (I)

D. (A) - (III), (B) - (I), (C) - (II), (D) - (IV)

Correct Option is (A)

AC generator	$\to$	Converts mechanical energy into electrical energy
Galvanometer	$\to$	Detects the presence of current in the circuit
Transformer	$\to$	Change AC voltage for smaller or greater value
Metal detector	$\to$	Works on the principle of resonance in AC circuit

7. (JEE Main 2021 (Online) 16th March Evening Shift)

For the given circuit, comment on the type of transformer used.

JEE Main 2021 (Online) 16th March Evening Shift Physics - Alternating Current Question 67 English

A. Auxiliary transformer

B. Step down transformer

C. Step-up transformer

D. Auto transformer

Correct Option is (C)

Voltage across secondary coil,

= $\frac{P o w e r a c r o s s l o a d}{C u r r e n t p a s \sin g t h r o u g h l o a d}$

$\Rightarrow V_{2} = \frac{P}{I_{L}} = \frac{60}{0.11} \Rightarrow V_{2} = 545.45$ V

Voltage across primary coil, V₁ = 220V

$\Rightarrow$ V₂ > V₁

It means that step-up transformer is used.

8. (JEE Main 2021 (Online) 24th February Morning Shift)

A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are ___________.

Correct Answer is (440)

In a transformer,

$\frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}}$

where, N_p = number of turns in primary circuit, N_s = number of turns in secondary circuit = 24, V_p = potential of primary circuit = 220 V and V_s = potential of secondary circuit = 12 V

$\Rightarrow \frac{N_{p}}{24} = \frac{220}{12}$

$\Rightarrow N_{p} = 440$

9. (JEE Main 2019 (Online) 9th January Evening Slot)

A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :

A. 50A

B. 45A

C. 35A

D. 25A

Correct Option is (B)

Given,

Primary voltage (V_P) = 2300 V

Primary current (I_P) = 5A

Secondary voltage (V_S) = 230 V

efficiency ( $η$ ) = 90%

We know,

Efficiency ( $η$ ) = $\frac{Sec o n d a r y P o w e r}{Pr i m a r y P o w e r}$

$∴$     $η$ = $\frac{P_{S}}{P_{P}}$ = $\frac{V_{S} I_{S}}{V_{P} I_{P}}$

$\Rightarrow$    0.9 = $\frac{230 \times I_{S}}{2300 \times 5}$

$\Rightarrow$    I = 45 A

10. (AIEEE 2003)

The core of any transformer is laminated so as to

A. reduce the energy loss due to eddy currents

B. make it light weight

C. make it robust and strong

D. increase the secondary voltage

Correct Option is (A)

Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.