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

Ques.150. Using Millman’s theorem, find the current through the load resistance RL of 3Ω resistance shown below (SSC-2014, M-Shift)

1. 12A
2. 4 A
3. 6A
4. 8A

Explanation:-

According to Millman’s theorem

Eeq = ∑EiYi/∑Yi

I1 = 8/3

I2 =16/3

I3 = 24/3 = 8A

Eeq = (8/3 + 16/3 + 8)/ (1/3 + 1/3 + 1/3) =16 V

Zeq = 1/∑Yi

= 1/(1/3 + 1/3 + 1/3) = 1Ω

Now the circuit become

I = 16/4 = 4 A

Ques.151. A stove element draws 15A when connected to 230 V line. How long does it take to consume one unit of energy? (SSC-2014, E-shift)

1. 3.45 h
2. 2.16 h
3. 1.0 h
4. 0.29 h

Explanation:-

P = E/T

Where E is the energy and T is time

T = E/P

1 Unit of energy is given as 1 Kwhr = 1000 Watt-hour

T = 1000/230 x 15    [since P = VI]

T = 0.29 hour

Ques.152. The Req for the circuit shown in figure is (SSC-2014, E-shift)

1. 14.4 Ω
2. 14.57 Ω
3. 15.27 Ω
4. 15.88 Ω

Explanation:-

In the above figure 6Ω and 3Ω are parallel therefore

(6 x 3)(6 + 3) = 2Ω

Now 2Ω and 2Ω, as well as 1Ω and 5Ω, are in series therefore

2Ω + 2Ω = 4Ω

5Ω + 1Ω = 6Ω

Ques.153. Calculate the voltage drop across 14.5 Ω  Resistance (SSC-2014, E-shift)

1. 14.5 V
2. 18 V
3. 29 V
4. 30.5 V

Explanation:-

In the given figure the resistances are in series, so the same current I will flow across the three resistors.

Requ = 14.5Ω + 25.5Ω + 60Ω = 100Ω

Current I = V/R

= 200/100 = 2A

Voltage drop across the 14.5Ω resistance

V = IR

= 2 x 14.5 = 29 V

Ques.154. For the network shown in the figure, the value of current in 8Ω resistors is (SSC-2014, E-shift)

1. 4. 8 A
2. 2.4 A
3. 1.5 A
4. 1.2 A

Explanation:-

The resistance between the point A and C will be

RAB || (RAC + RBC)

AB =  20 x(12 + 8)/(20 + 12 + 8) = 10Ω

I = V/R = 48/10 = 4.8 A

Since AC and BC are parallel with AB, therefore, the current will divide equally into both side

hence the current through the Branch ABC is = I/2

= 4.8/2 = 2.4 A

Ques.155. A piece of oil-soaked paper has been inserted between the plates of a parallel plate capacitor. Then the potential difference between the plates will (SSC-2014, E-shift)

1. Increase
2. Decrease
3. Remain Unaltered
4. Become Zero

Explanation:-

The paper sheet is the poor conductor of electricity so it does not allow the flow of electric current or electric charges between two parallel plates. However, the paper sheet allows the electric field through it. Therefore, the paper sheet placed between the parallel plates acts as the barrier for the electric current. Hence the potential difference between the plates will decrease.

Ques.156. Tesla is same as (SSC-2014, E-shift)

1. Weber/meter
2. Weber/(meter)2
3. Greater than
4. Double

Explanation:-

The Tesla (symbol T) is the SI derived unit used to measure magnetic fields. One tesla is equal to one weber per square meter.

Ques.157. The unit of volume of resistivity is (SSC-2014, E-shift)

1. Ohm-m3/m2
2. Ohm-m2/m
3. Ohm-gram-m/gram
4. Ohm-m4/m3

Explanation:-

Resistivity is the resistance of a unit volume of a material. In the metric system, the unit of length is the meter, and the area is the square meter. Thus, resistivity is measured in units of Ohm – meters squared per meter (Ohm-m2/m), often abbreviated as Ohm-m.

Ques.158. Four resistance 2Ω, 4Ω, 5Ω, 20Ω are connected in parallel. Their combined resistance is (SSC-2014, E-shift)

Since the resistance is connected in parallel therefore

1/ Req = 1/2Ω + 1/4Ω + 1/5Ω + 1/20Ω

=10 + 5 + 4 + 1/20

Req= 1Ω

Note:- The total equivalent resistor connected in parallel is always less than the smallest given resistance. Here in the question, the smallest resistance is 2Ω and in the given option only option 1 i.e 1 Ω is less than the smallest resistance

so in the above case, you can directly find the answer without any calculation.

Ques.159. In the given figure, the Value of R is (SSC-2014, E-shift)

1. 2.5Ω
2. 7.5Ω
3. 10Ω

Explanation

Here 10Ω resistance is connected in parallel

R’ = 10 x10/10 + 10 = 5Ω

Req = R + R’ = (R + 5)Ω

V = IR

100 =  8(R + 5)

100 -40 = 8R

R = 7.5Ω

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