1

### JEE Main 2018 (Online) 16th April Morning Slot

A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ;
A
50 A
B
45 A
C
25 A
D
20 A

## Explanation

Efficiency n = 0.9 = ${{{P_s}} \over {{P_p}}}$

as $\,\,\,\,\,\,$ P = VI

$\therefore\,\,\,$ Ps = 0.9 $\times$ Pp

$\Rightarrow$ $\,\,\,\,$ Vs Is = 0.9 $\times$ Vp Ip

$\Rightarrow$ $\,\,\,\,$ Is = ${{0.9 \times 2300 \times 5} \over {230}}$

= $\,\,\,\,$ 45 A
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### JEE Main 2018 (Online) 16th April Morning Slot

A plane electromagnetic wave of wavelength $\lambda$ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by :
A B C D 3

### 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
50 A
B
45 A
C
35 A
D
25 A

## Explanation

Given,

Primary voltage (VP) = 2300 V

Primary current (IP) = 5A

Secondary voltage (VS) = 230 V

efficiency ($\eta$) = 90%

We know,

Efficiency ($\eta$) = ${{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}$

$\therefore$   $\eta$ = ${{{P_S}} \over {{P_P}}}$ = ${{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}$

$\Rightarrow$   0.9 = ${{230 \times {{\rm I}_S}} \over {2300 \times 5}}$

$\Rightarrow$   I = 45 A
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### JEE Main 2019 (Online) 10th January Morning Slot

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin $\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$ then the maximum electric field associated with it is -
A
4.5 $\times$ 104 N/C
B
4 $\times$ 104 N/C
C
6 $\times$ 104 N/C
D
3 $\times$ 104 N/C

## Explanation

E0 = B0 $\times$ C

= 100 $\times$ 10$-$6 $\times$ 3 $\times$ 108

= 3 $\times$ 104 N/C

$\therefore$  correct answer is 3 $\times$ 104 N/C