JEE Main Alternating Current PYQ

1.(JEE Main 2024 (Online) 30th January Morning Shift )

Primary coil of a transformer is connected to $220 V$ ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistances shown in figure. The output voltage $(V_{0})$ is :

JEE Main 2024 (Online) 30th January Morning Shift Physics - Alternating Current Question 16 English

A.7 V

B.44 V

C.22 V

D.15 V

Correct option is (a)

$\begin{aligned} \frac{ε_{1}}{ε_{2}} = \frac{N_{1}}{N_{2}} = \frac{100}{10} \Rightarrow ε_{2} = 22 V \\ I = \frac{22}{22 \times 10^{3}} = 1 mA, V_{0} = 7 V \end{aligned}$

2.(JEE Main 2024 (Online) 27th January Evening Shift )

Primary side of a transformer is connected to $230 V, 50 Hz$ supply. Turns ratio of primary to secondary winding is $10 : 1$ . Load resistance connected to secondary side is $46 Ω$ . The power consumed in it is :

A.11.5 W

B.12.5 W

C.10.0 W

D.12.0 W

Correct option is (a)

$\begin{aligned} \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}} \\ \frac{230}{V_{2}} = \frac{10}{1} \\ V_{2} = 23 V \end{aligned}$

Power consumed $= \frac{V_{2}^{2}}{R}$

$= \frac{23 \times 23}{46} = 11.5 W$

3.(JEE Main 2024 (Online) 30th January Evening Shift )

A power transmission line feeds input power at $2.3 kV$ to a step down transformer with its primary winding having 3000 turns. The output power is delivered at $230 V$ by the transformer. The current in the primary of the transformer is $5 A$ and its efficiency is $90 %$ . The winding of transformer is made of copper. The output current of transformer is _________ $A$ .

Correct answer is 45

$\begin{aligned} P_{i} = 2300 \times 5 watt \\ P_{0} = 2300 \times 5 \times 0.9 = 230 \times I_{2} \\ I_{2} = 45 A \end{aligned}$

4. (JEE Main 2023 (Online) 6th April Morning Shift)

An ideal transformer with purely resistive load operates at $12 kV$ on the primary side. It supplies electrical energy to a number of nearby houses at $120 V$ . The average rate of energy consumption in the houses served by the transformer is 60 $kW$ . The value of resistive load $(Rs)$ required in the secondary circuit will be ___________ $m Ω$ .

Correct Answer is (240)

The power delivered to the houses is given as 60 kW. This power is supplied at a voltage of 120 V. The power consumed in a resistive load can be found using the formula $P = V^{2} / R$ , where P is the power, V is the voltage, and R is the resistance.

We can rearrange this formula to solve for the resistance:

$R = V^{2} / P$

Substituting the given values gives:

$R = (120 V)^{2} / 60, 000 W = 0.24 Ω$

Since we want the resistance in milliohms (mΩ), we can convert this to milliohms by multiplying by 1000:

$R = 0.24 Ω \times 1000 = 240 m Ω$

So, the value of resistive load required in the secondary circuit is 240 mΩ.

5. (JEE Main 2022 (Online) 28th July Evening Shift)

A transformer operating at primary voltage $8 kV$ and secondary voltage $160 V$ serves a load of $80 kW$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

A. $800 Ω$ and $1.06 Ω$

B. $10 Ω$ and $500 Ω$

C. $800 Ω$ and $0.32 Ω$

D. $1.06 Ω$ and $500 Ω$

Correct Option is (C)

$V_{1} i_{1} = V_{2} i_{2} = 80$ kW

$\Rightarrow i_{1} = 10 A$ and $i_{2} = \frac{80 \times 1000}{160} = 500 A$

$\Rightarrow R_{1} = \frac{V_{1}}{i_{1}} = 800 Ω$ and $R_{2} = \frac{160}{500} = 0.32 Ω$