SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-2)

Ques.120. In a parallel RLC circuit if the lower cut-off frequency is 2400 Hz and the uppercut off frequency is 2800 Hz, What is the bandwidth? (SSC-2015)

1. 2800 Hz
2. 2400 Hz
3. 400 Hz
4. 5200 Hz

Explanation:-

Bandwidth = upper frequency – lower frequency

Bandwidth = f1 – f2

2800 – 2400

400 Hz.

Ques.121. Two wires A and B have the same cross-section and are made of the same material. RA = 800Ω and RB = 100Ω. The number of times A is longer than B is (SSC-2014, M-Shift)

1. 5
2. 6
3. 2
4. 4

None of the options is correct

Explanation:-

Since they are of the same material they will have the same resistivity.

Therefore their length now

Ra/Rb=la/lb

800/100 = la/lb

la = 8lb

Therefore A is 8 times longer than B

Ques.122. In the circuit shown in the figure, find the transient current i(t) when the switch is closed at t = 0, Assume zero initial condition.

1. 50 t e-0.5t
2. 50 t e-5t
3. 100 t e-5t
4. 100 t e-0.5t

Explanation:-

Ques.123. The power factor of the circuit shown in figure: (SSC-2014, M-Shift)

1. 0.75 lagging
2. 0.6 lagging
3. 0.3 lagging
4. 0.8 lagging

Explanation:-

The power factor of series RL circuit

Due to the inductive circuit power factor is lagging.

Ques.124. The power factor of the A.C circuit is given by: (SSC-2014, M-Shift)

1. R/Z
2. XL/R
3. Z/R
4. R/XL

Explanation:-

In electrical engineering, the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit ” R/Z”.

Ques.125. Two 100 W, 200 V lamps are connected in series across a 200 V supply. The total power consumed by each lamp will be watts: (SSC-2014, M-Shift)

1. 200
2. 25
3. 50
4. 100

Explanation:-

Let’s take a 200 volts – 100  watts bulb and calculate the resistance of its filament :

We know,

P=V2R

or

R=V2P

When V=200 volts and P = 100 watts,

R = (200*200)/100 = 400 ohms.

Since now the two bulbs are in series, the total resistance offered by them is :

R’ = 400 Ω + 400 Ω = 800 Ω

Now the current that flows through either of them is :

I=V/R

=200/800=0.25A

Now total power consumed by each lamp

P=I2R

P=(0.25 x 0.25)400

P = 25 watts

Ques.126. The Bio-Savart’s law is a general modification of: (SSC-2014, M-Shift)

2. Kirchhoff’s law
3. Lenze law
4. Ampere’s law

Explanation:-

Biot-Savart’s law describes the three different types of currents, which is nothing but the filament or line current, surface current and volume current. Hence, this law is a general modification of Ampere’s circuital law.

Ques.127. The active and reactive power of an inductive circuit are 60 W and 80 VAR respectively. The power factor of the circuit is (SSC-2014, M-Shift)

1. 0.8 lag
2. 0.5 lag
3. 0.6 lag
4. 0.75 lag

Explanation:-

Active Power P = 60 W

Reactive Power Q = 80 VAR

S =√P2 + Q2 = √602 + 802

= 100 VA

Power factor = cos Φ = P/S = 60/100 =0.6

Second Method

Ques.128. Which of the following is the non-linear circuit parameter? (SSC-2014, M-Shift)

1. Transistor
2. Inductance
3. Condenser
4. Wire Wound Resistor

Explanation:-

A linear circuit is an electric circuit in which circuit parameters (Resistance, inductance, capacitance, waveform, frequency, etc) are constant. In other words, a circuit whose parameters are not changed with respect to Current and Voltage is called Linear Circuit.

Whereas in a non-linear circuit the parameters (Resistance, inductance, capacitance, etc) is not constant, which is called Non-Linear Circuit.

Example : – Diode,transistor,SCR , transformer, iron core, inductor etc.

Ques.129. A terminal where three or more branches meet is known as (SSC-2014, M-Shift)

1. Mesh
2. Node
3. Terminus
4. Loop