Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

In a series $$LCR$$ circuit $$R = 200\Omega $$ and the voltage and the frequency of the main supply is $$220V$$ and $$50$$ $$Hz$$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $${30^ \circ }.$$ On taking out the inductor from the circuit the current leads the voltage by $${30^ \circ }.$$ The power dissipated in the $$LCR$$ circuit is

A

$$305$$ $$W$$

B

$$210$$ $$W$$

C

$$zero$$ $$W$$

D

$$242$$ $$W$$

When capacitance is taken out, the circular is $$LR.$$

$$\therefore$$ $$\tan \phi = {{\omega L} \over R}$$

$$ \Rightarrow \omega L = R\,\tan \phi $$

$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$

Again, when inductor is taken out, the circuit is $$CR.$$

$$\therefore$$ $$\tan \phi = {1 \over {\omega CR}}$$

$$ \Rightarrow {1 \over {\omega c}} = R\tan \phi $$

$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$

Now, $$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}} $$

$$ = \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega $$

Power dissipated $$ = {V_{rms}}{I_{rms}}\cos \phi $$

$$ = {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$$

$$\left( {\,\,\,} \right.$$ as $$\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$$

$$ = {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$$

$$ = {{220 \times 220} \over {200}} = 242\,W$$

$$\therefore$$ $$\tan \phi = {{\omega L} \over R}$$

$$ \Rightarrow \omega L = R\,\tan \phi $$

$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$

Again, when inductor is taken out, the circuit is $$CR.$$

$$\therefore$$ $$\tan \phi = {1 \over {\omega CR}}$$

$$ \Rightarrow {1 \over {\omega c}} = R\tan \phi $$

$$ = 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$$

Now, $$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}} $$

$$ = \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega $$

Power dissipated $$ = {V_{rms}}{I_{rms}}\cos \phi $$

$$ = {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$$

$$\left( {\,\,\,} \right.$$ as $$\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$$

$$ = {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$$

$$ = {{220 \times 220} \over {200}} = 242\,W$$

2

MCQ (Single Correct Answer)

An inductor of inductance $$L=400$$ $$mH$$ and resistors of resistance $${R_1} = 2\Omega $$ and $${R_2} = 2\Omega $$ are connected to a battery of $$emf$$ $$12$$ $$V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t=0.$$ The potential drop across $$L$$ as a function of time is :

A

$${{12} \over t}{e^{ - 3t}}V$$

B

$$6\left( {1 - {e^{ - t/0.2}}} \right)V$$

C

$$12{e^{ - 5t}}V$$

D

$$6{e^{ - 5t}}V$$

Growth in current in $$L{R_2}$$ branch when switch is closed is given by

$$i = {E \over {{R_2}}}\left[ {1 - {e^{ - {R_2}t/L}}} \right]$$

$$ \Rightarrow {{di} \over {dt}} = {E \over {{R_2}}}.{{{R_2}} \over L}.{e^{ - {R_{2t/L}}}}$$

$$ = {E \over L}{e^{{{{R_2}t} \over L}}}$$

Hence, potential drop across

$$L = \left( {{E \over L}{e^{ - {R_2}t/L}}} \right)L = E{e^{ - {R_2}t/L}}$$

$$ = 12{e^{ - {{2t} \over {400 \times {{10}^{ - 3}}}}}} = 12{e^{ - 5t}}V$$

$$i = {E \over {{R_2}}}\left[ {1 - {e^{ - {R_2}t/L}}} \right]$$

$$ \Rightarrow {{di} \over {dt}} = {E \over {{R_2}}}.{{{R_2}} \over L}.{e^{ - {R_{2t/L}}}}$$

$$ = {E \over L}{e^{{{{R_2}t} \over L}}}$$

Hence, potential drop across

$$L = \left( {{E \over L}{e^{ - {R_2}t/L}}} \right)L = E{e^{ - {R_2}t/L}}$$

$$ = 12{e^{ - {{2t} \over {400 \times {{10}^{ - 3}}}}}} = 12{e^{ - 5t}}V$$

3

MCQ (Single Correct Answer)

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $$A=$$ $$10\,\,c{m^2}$$ and length $$=20$$ $$cm$$ . If one of the solenoid has $$300$$ turns and the other $$400$$ turns, their mutual inductance is

$$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,Tm\,{A^{ - 1}}} \right)$$

$$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,Tm\,{A^{ - 1}}} \right)$$

A

$$2.4\pi \times {10^{ - 5}}H$$

B

$$4.8\pi \times {10^{ - 4}}H$$

C

$$4.8\pi \times {10^{ - 5}}H$$

D

$$2.4\pi \times {10^{ - 4}}H$$

$$M = {{{\mu _0}{N_1}{N_2}A} \over \ell }$$

$$ = {{4\pi \times {{10}^{ - 7}} \times 300 \times 400 \times 100 \times {{10}^{ - 4}}} \over {0.2}}$$

$$ = 2.4\pi \times {10^{ - 4}}H$$

$$ = {{4\pi \times {{10}^{ - 7}} \times 300 \times 400 \times 100 \times {{10}^{ - 4}}} \over {0.2}}$$

$$ = 2.4\pi \times {10^{ - 4}}H$$

4

MCQ (Single Correct Answer)

An ideal coil of $$10H$$ is connected in series with a resistance of $$5\Omega $$ and a battery of $$5V$$. $$2$$ second after the connection is made, the current flowing in ampere in the circuit is

A

$$\left( {1 - {e^{ - 1}}} \right)$$

B

$$\left( {1 - e} \right)$$

C

$$e$$

D

$${{e^{ - 1}}}$$

(When current is in growth in $$LR$$ circuit)

$$ = {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)$$

$$ = {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)$$

$$ = \left( {1 - {e^{ - 1}}} \right)$$

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