1

### JEE Main 2017 (Online) 9th April Morning Slot

A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 $\Omega$, L=25 mH and C=1000 $\mu$F. The total impedance, and phase difference between the voltage across the source and the current will respectively be :
A
10 $\Omega$ and tan$-$1 $\left( {{5 \over 3}} \right)$
B
$7\,\Omega$ and 45o
C
$10\,\Omega$ and tan$-$1$\left( {{8 \over 3}} \right)$
D
$7\,\Omega$ and tan$-$1$\left( {{5 \over 3}} \right)$
2

### JEE Main 2017 (Online) 9th April Morning Slot

The electric field component of a monochromatic radiation is given by

$\overrightarrow E$ = 2 E0 $\widehat i$ cos kz cos $\omega$t

Its magnetic field $\overrightarrow B$ is then given by :
A
${{2{E_0}} \over c}$ $\widehat j$ sin kz cos $\omega$t
B
$-$ ${{2{E_0}} \over c}$ $\widehat j$ sin kz sin $\omega$t
C
${{2{E_0}} \over c}$ $\widehat j$ sin kz sin $\omega$t
D
${{2{E_0}} \over c}$ $\widehat j$ cos kz cos $\omega$t
3

### JEE Main 2018 (Offline)

In an a.c. circuit, the instantaneous e.m.f. and current are given by
e = 100 sin 30 t
i = 20 sin $\left( {30t - {\pi \over 4}} \right)$
In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively
A
50, 0
B
50, 10
C
${{1000} \over {\sqrt 2 }},10$
D
${{50} \over {\sqrt 2 }}$

## Explanation

Wattless current,

here  $\phi$  is the angle between i and e.

Average power,

Pav = Vrms Irms cos$\phi$

= ${{100} \over {\sqrt 2 }} \times {{20} \over {\sqrt 2 }}$ cos${\pi \over 4}$

= ${{1000} \over {\sqrt 2 }}$ watt.
4

### JEE Main 2018 (Offline)

For an RLC circuit driven with voltage of amplitude vm and frequency ${\omega _0}$ = ${1 \over {\sqrt {LC} }}$ the current exhibits resonance. The quality factor, Q is given by :
A
${{CR} \over {{\omega _0}}}$
B
${{{\omega _0}L} \over R}$
C
${{{\omega _0}R} \over L}$
D
${R \over {\left( {{\omega _0}C} \right)}}$

## Explanation

Quality factor (Q) = ${{Angular\,\,{\mathop{\rm Re}\nolimits} sonance} \over {Bandwith}}$

= ${{{1 \over {\sqrt {LC} }}} \over {{R \over L}}}$

= ${{{\omega _0}} \over {{R \over L}}}$

= ${{{\omega _0}L} \over R}$

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