# GATE EE 2004

 Question 1
The value of Z in figure, which is most appropriate to cause parallel resonance at 500 Hz is
 A 125.00 mH B 304.20 $\mu F$ C 2.0 $\mu F$ D 0.05 $\mu F$
Electric Circuits   Resonance and Locus Diagrams
Question 1 Explanation:
Atresonance, the circuit should be in unity power factor.
Hence, 'Z' should be capacitive.
Admittance of the parallel circuit.
$Y=\frac{1}{jL\omega }+\frac{1}{1/jC\omega }=0$
$\frac{-1}{L\omega }+C\omega =0$
$\therefore \;\; C=\frac{1}{L\omega ^2}=\frac{1}{2(2 \pi \times 500)^2} =0.05\mu F$
 Question 2
A parallel plate capacitor is shown in figure. It is made two square metal plates of 400 mm side. The 14 mm space between the plates is filled with two layers of dielectrics of $\varepsilon _r$= 4, 6 mm thick and $\varepsilon _r$= 2, 8 mm thick. Neglecting fringing of fields at the edge the capacitance is
 A 1298 pF B 944 pF C 354 pF D 257pF
Electromagnetic Fields   Electrostatic Fields
Question 2 Explanation:
When two capacitor formed by two layer of dielectics are connected in series, then equivalent capacitance
\begin{aligned} C_{eq} &= \frac{C_1 \times C_2}{C_1+C_2}\\ &=\frac{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} \times \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}}{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} + \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}} \\ &= \frac{(8.85 \times 10^{-12})(400 \times 10^{-3})^2 \times 4 \times 2}{4 \times 8 \times 10^{-3}+6 \times 2 \times 10^{-3}} \\ & =257 \; \text{pF} \end{aligned}

 Question 3
The inductance of a long solenoid of length 1000 mm wound uniformly with 3000 turns on a cylindrical paper tube of 60 mm diameter is
 A 3.2 $\mu H$ B 3.2 mH C 32.0 mH D 3.2 H
Electromagnetic Fields   Magnetostatic Fields
Question 3 Explanation:
Let, $B=\mu _0nI$
Then, $\phi =B.S=\mu _0nSI$
Where 'S' is the cross sectional area of solenoid flux linkage,
$a'=n\phi =\mu _0n^2SI$
Hence, Inductance length $=\mu _0n^2S$
For,
$l=1000mm=1m$
$\therefore \;\;L=4 \pi \times 10^{-7} \times 3000^2 \times \pi \times (30 \times 10^{-3})^2$
$\;\;\;=32mH$
 Question 4
Total instantaneous power supplied by a 3-phase ac supply to a balanced R-L load is
 A zero B constant C pulsating with zero average D pulsating with the non-zero average
Power Systems   Power System Transients
Question 4 Explanation:
Impedance of the load
$=Z_L=R+j\omega L=|Z|\angle \theta _L$
where, $\theta _L=\tan ^{-1}\left ( \frac{\omega L}{R} \right )$
Voltages of $3-\phi$ supply
\begin{aligned} V_a&=V_m \sin \omega t \\ V_b&=V_m \sin (\omega t-120^{\circ}) \\ V_c &=V_m \sin (\omega t+120^{\circ}) \\ I_a&=\frac{V_a}{Z_L}=\frac{V_m \sin \omega t}{|Z|\angle \theta _L} \\ &= I_M \sin (\omega t-\theta _L)\\ \text{where, } I_m&=\frac{V_m}{|Z|}\\ \text{Similarly,}&\\ I_b&=I_m \sin (\omega t-120-\theta _L)\\ I_c&=I_m \sin (\omega t+120-\theta _L)\\ & \text{Instantaneous power}\\ &=P=V_aI_a+V_bI_b+V_cI_c\\ P&=V_mI_m[\sin \omega t\cdot \sin (\omega t-\theta _L)\\ &+\sin (\omega t-120^{\circ}) \cdot \sin (\omega t-120-\theta _L)\\ &+\sin (\omega t+120^{\circ})\cdot \sin (\omega t+120-\theta _L)]\\ &=\frac{V_mI_m}{2}[(\cos \theta _L -\cos(2\omega t-\theta _L) ) \\ &+(\cos \theta _L -\cos(2\omega t-240-\theta _L) )\\ &+(\cos \theta _L -\cos(2\omega t+240-\theta _L) )]\\ P&=\frac{3V_mI_m}{2}\cos \phi = \text{constant} \end{aligned}
 Question 5
A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is
 A 50.00% B 70.70% C 141.40% D 200.00%
Electrical Machines   Transformers
Question 5 Explanation:
Copper loass at any load
\begin{aligned} P_{cu} &=x^2 \times [\text{Full load copper loss}] \\ &= x^2 \times P_{fl,cu}\\ \text{where,}\;\; x&= \text{fraction of laod} \end{aligned}
Maximum occurs when
Copper loss-Iron loss
$x^2 \times P_{fl,cu}=P_i$
$x^2 \times 600=300 x=0.707 =70.7\%$

There are 5 questions to complete.