Question 1 |
The value of Z in figure, which is most appropriate to cause parallel resonance
at 500 Hz is
125.00 mH | |
304.20 \mu F | |
2.0 \mu F | |
0.05 \mu F |
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
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
1298 pF | |
944 pF | |
354 pF | |
257pF |
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}
\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
3.2 \mu H | |
3.2 mH | |
32.0 mH | |
3.2 H |
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
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
zero | |
constant | |
pulsating with zero average | |
pulsating with the non-zero average |
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}
=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
50.00% | |
70.70% | |
141.40% | |
200.00% |
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\%
\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.