# SSC JE Electrical Conventional Paper Solved 2014-Electrical-Exam

# SSC JEÂ Electrical Conventional Paper with Explained SolutionÂ 2014 | MES Electrical

**Ques 1(a):-Â An oven operated on a 15.0 A current from a 120 V source. How much energy will it consume in 3.0 h of operation?**

**Answer:-Â **Energy consumption or power consumption refers to the electrical energy per unit time is given as

E = PowerÂ Ã— Time = PÂ Ã— T

Given:-

Source Voltage = 120 V

Current = 15 A

Time = 3 hours

Power = VI

P = 120Â Ã— 15 = 1800 watts

**Energy consumption in 3 Hours**

**= 1800Â Ã— 3 = 5400 Whr = 5.4 kWh**

**Ques 1(b).Â How many 100 W light bulbs connected to a 120 V supply can be turned on at the same time without blowing a 15.0 A fuse?**

**Answer:-Â Â **GivenÂ

Supply voltage V = 120

Power of Bulb P = 100 W

Current I = 15 A

Now the power in an A.C circuit is given as

P = VI

P = 120Â Ã— 15 = 1800 Watts

Number of Bulbs = Power Available/Power Per Bulb

**N = P/P _{n} = 1800/100 = 18**

**Hence we can illuminate 18 Bulbs at the same time without blowing 15 A fuse.**

**Ques 1 (c). A 120 V circuit containsÂ 10Î©. What resistance must be added in series for the circuit to have a current of 5.0 A?**

**Answer:-Â **Given

Supply Voltage V = 120 V

ResistanceÂ = 10 ohm

Require current = 5 A

Let “R” Be the unknown resistanceÂ to be added in the series then by ohms law

**V = IR**

120 = 5 Ã— (10 + R)

24 = 10 + R

R = 24 âˆ’ 10

**R = 14 ohm**

**Hence 14 Î© resistance must be added in series for the circuit to have a current of 5.0 A.**

**Ques 1(d).**Â **In the following circuit, find the total resistance R _{3}, V_{2}, and I_{4}.**

**Answer:-Â ** Given

R_{1} = 9Î©, R_{2} = 4Î©

R_{4} = 12Î© , R_{5} = 36Î©

V_{t} = 12 V, I_{t} = 1 A

R_{3} = ?, V_{2} = ?, I_{4} = ?

The given circuit can be reconstructed as shown in the figure

The Resistance **R _{1} and R_{3}** are in

**series**therefore equivalent Resistance is

**R _{eq1} = R_{1} + R_{3} =**

**(9 + R3)Î©**

Resistance **R _{4}** and

**R**are inÂ parallel hence equivalent resistance

_{5}**R _{eq2} = (R_{4}Â Ã— R_{5})/(R_{4} + R_{5})**

= (12Â Ã— 36)/(12 + 36) = 432/48 = **9Î©**

After solving the resistance parameter the equivalent circuit can be reduced as shown in the figure

Total equvalent resistance of the circuit R_{eq} = (R_{eq1} || 4) + R_{eq2}

**Req = [(9 + R3) || 4] + 9 ——————(1)**

Also R_{t} = V_{t}/I_{t} = 12/1 = **12Î©—————(2)**

Equating equation 1 and 2 we get

**â‡’ Voltage across Resistance R _{2}**

Now resistance (R_{1} + R_{3}) and R_{2} are in parallel therefore by using current divison rule

I_{2} = I_{R2}

Hence the voltage across resistance R2

**V _{2} = I_{2}(R_{2}) = 0.75Â Ã— 9 = 3V**

or

**V _{2} = V_{equ} = R_{3}Â Ã— I’_{equ} = 3Â Ã— 1 = 3V**

**â‡’ Calculation of Current I _{4}**

Since Resistance R4 and R5 in parallel

So using current division rule

I_{4} = I_{R4}

So R_{3} = 3Î©

V_{2} = 3V

**I _{4} = 0.75 A**

or

V_{4} = V_{t}Â âˆ’ V_{2}

= 12Â âˆ’ 3 = 9V

**I _{4} = V_{4}/R_{4} = 9/12 = 0.75 A**

Therefore

**R _{3} = 3Î©, V_{2} = 3V, I_{4} = 0.75A**