Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

Numerical

At very high frequencies, the effective impendence of the given circuit will be ________________ $$\Omega$$.

Your Input ________

Correct Answer is **2**

X_{L} = 2$$\pi$$fL

f is very large

$$\therefore$$ X_{L} is very large hence open circuit.

$${X_C} = {1 \over {2\pi fC}}$$

f is very large.

$$\therefore$$ X_{C} is very small, hence short circuit.

Final circuit

$$ \therefore $$ $${Z_{eq}} = 1 + {{2 \times 2} \over {2 + 2}} = 2$$

f is very large

$$\therefore$$ X

$${X_C} = {1 \over {2\pi fC}}$$

f is very large.

$$\therefore$$ X

Final circuit

$$ \therefore $$ $${Z_{eq}} = 1 + {{2 \times 2} \over {2 + 2}} = 2$$

2

Numerical

An ac circuit has an inductor and a resistor resistance R in series, such that X_{L} = 3R. Now, a capacitor is added in series such that X_{C} = 2R. The ratio of new power factor with the old power factor of the circuit is $$\sqrt 5 :x$$. The value of x is ___________.

Your Input ________

Correct Answer is **1**

$$\cos \phi = {R \over {\sqrt {{R^2} + 3{R^2}} }}$$

$$ = {1 \over {\sqrt {10} }}$$

$$\cos \phi ' = {R \over {\sqrt {{R^2} + {R^2}} }}$$

$$ = {1 \over {\sqrt 2 }}$$

$${{\cos \phi '} \over {\cos \phi }} = {{\sqrt {10} } \over {\sqrt 2 }} = {{\sqrt 5 } \over 1}$$

$$\therefore$$ x = 1

3

Numerical

The alternating current is given by $$i = \left\{ {\sqrt {42} \sin \left( {{{2\pi } \over T}t} \right) + 10} \right\}A$$

The r.m.s. value of of this current is ................. A.

The r.m.s. value of of this current is ................. A.

Your Input ________

Correct Answer is **11**

$$f_{rms}^2 = f_{1\,rms}^2 + f_{2\,rms}^2$$

$$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$$

$$ = 121 \Rightarrow {f_{rms}}$$ = 11 A

$$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$$

$$ = 121 \Rightarrow {f_{rms}}$$ = 11 A

4

Numerical

In the given figure the magnetic flux through the loop increases according to the relation $$\phi$$_{B}(t) = 10t^{2} + 20t, where $$\phi$$_{B} is in milliwebers and t is in seconds.

The magnitude of current through R = 2$$\Omega$$ resistor at t = 5 s is ___________ mA.

The magnitude of current through R = 2$$\Omega$$ resistor at t = 5 s is ___________ mA.

Your Input ________

Correct Answer is **60**

$$\left| \in \right| = {{d\phi } \over {dt}}$$ = 20t + 20 mV

$$\left| i \right| = {{\left| \in \right|} \over R}$$ = 10t + 10 mA

at t = 5

$$\left| i \right|$$ = 60 mA

$$\left| i \right| = {{\left| \in \right|} \over R}$$ = 10t + 10 mA

at t = 5

$$\left| i \right|$$ = 60 mA

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