Ques.71. If in an R-L-C series circuit the current lags the applied voltage by 60° then (SSC – 2011)
XL – Xc = R/√3
XL – Xc = √3R
XL = Xc = R
XL – Xc = R
Answer.2. XL – Xc = √3R
Explanation:-
Since Angle between V and I is 60°
Then Power factor angle = θ = 60°
For series RLC Circuit the phase difference between the current and the voltage is
tanθ = (XL – Xc)/R ……. (Since XL > Xc)
tan60° = (XL – Xc)/R
√3R = (XL – Xc)
Ques.72. A lossy capacitor with a loss angle of 0.01 radian, draws a current of 0.5 A when supplied at 1000 V from a sinusoidal voltage source. The active power consumed by the capacitor is (SSC – 2011)
5 W
10 W
2 W
1 W
Answer.1. 5 W
Explanation:-
Loss angle = 0.01 radian
θL = 0.01 x 180/π = 0.57 ———- (Converting radian into degree)
P.F. angle = 90 – 0.57 = 89.43
Active power consumed = V I cosθ
= 1000 × 0.5 × cos(89.427)
= 4.999 ≅ 5
Ques.73. An AC voltage source with an internal impedance Z1 is connected to a load of impedance Z2. For maximum power transfer to the load, the condition is (SSC – 2011)
Z2 = Z1
|Z2| =|Z1|
Z2* = Z1
Z2 = Z1*
Answer.4. Z2 = Z1*
Explanation:-
For maximum power transfer, input Impedance must be equal to the conjugate of output Impedance.
As we know that the impedance of input is something to which a signal is applied to measure of how much power that input will tend to draw (from a given output voltage). This impedance is known as the load impedance.
Therefore Z2 = Z1*
Ques.74. A current wave starts at zero, rises instantaneously, then remains at a value of 20 A for 10 sec, then decreases instantaneously, remaining at a value of – 10 A for 20 sec, and then repeats this cycle. The RMS value of the wave is (SSC – 2011)
22.36A
17.32 A
8.165 A
14.14 A
Answer.4. 14.14 A
Explanation:-
RMS Value of the current of the sine wave is
Ques.75. Two incandescent bulbs of rating 230 V, 100 W, and 230 V, 500 W are connected in parallel across the mains. As a result what will happen? (SSC – 2011)
100 W bulb will glow brighter
500 W bulb will glow brighter
Both the bulbs will glow equally bright
Both the bulbs will glow dimly
Answer.2. 500 W bulb will glow brighter
Explanation:-
In a parallel connection, the voltage across each element is the same. So when 100W bulb and 500W bulb are connected in parallel, the voltage across them will be the same (230 V in the given case). To find which bulb will glow brighter we need to find the power dissipation across each of them. From the relation
P = V2/R
R100 = 2302/100 = 529 Ohm
R500 = 2302/500 = 105.8 Ohm
Since the voltage is the same we can say that power dissipation will be higher for the bulb with lower resistance i.e. 500W bulb.
Therefore 500-watt bulb will glow brighter.
Ques.76. The voltage wave given by V = 4cosωt produces a current wave i = 1.5cosωt – 2.598sinωt in a circuit. The current wave (SSC – 2011)
Leads voltage wave by 60°
Lags voltage wave by 60°
Leads voltage wave by 30°
Lags voltage wave by 30°
Answer.1. Leads voltage wave by 60°
Explanation:-
Given,
V = 4cosωt
i = 1.5cosωt – 2.598sinωt
= 1.5cosωt + 2.598sin(ωt + π/2)
Or I = Ia(cosωt + θ)……………(sinusoidal current wave)
Where Ia is amplitude or peak value of current which is given as
Ques.77. An electric load consumed 17.32 kW at a power factor of 0.707 (lagging). For changing the load power factor to 0.866 (lagging), the capacitor that is to be connected in parallel with the load should draw (SSC – 2011)
7.32 kVAR
10 kVAR
27.32 kVAR
10.32 kVAR
Answer.1. 7.32 kVAR
Explanation:-
Leading kVAr taken by the capacitor,
Q = P(tanΦ1 – tanΦ2)
CosΦ1 = 0.707
Φ1 = 45°
CosΦ2 = 0.866
Φ2 = 30°
Q = 17.32 (tan 45° – tan 30°)
= 7.32 KVAR
Ques.78. The value of voltage across the diode in the figure given below is (SSC – 2010)
0 Volt
4 Volt
8 Volt
Depend upon the value of R
Answer.3. 8 Volt
Explanation:-
There will be no current in the circuit as the diode is reverse biased and the diode is reversed biased offer infinite resistance. You can think of a reverse-biased diode as an open circuit.
So, the whole voltage will across the diode i.e 8V
Ques.79. The ratio of resistance of a 100 W, 220 V lamp to that of a 100 W, 110 V lamp will be at respective voltages (SSC – 2010)
4
2
1/2
1/4
Answer.1. 4
Explanation:-
For 100 W, 220 V
R1 = V2/P = 2202/100 = 484
For 100 W, 110 V
R2 = V2/P = 1102/100 = 121
Ratio of resistance R1 and R2
R1/R2 = 484/121 = 4
Ques.80. Two sinusoidal equations are given as (SSC – 2010)
E1 = A sin(ωt + π/4)
E2 = B sin(ωt – π/6)
The phase difference between the two quantities is:
