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

# SSC JE Electrical Conventional Paper with Explained Solution 2014 | MES Electrical

**Ques 1(a):- 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:- **Energy consumption or power consumption refers to the electrical energy per unit time is given as

E = Power × Time = P × T

Given:-

Source Voltage = 120 V

Current = 15 A

Time = 3 hours

Power = VI

P = 120 × 15 = 1800 watts

**Energy consumption in 3 Hours**

**= 1800 × 3 = 5400 Whr = 5.4 kWh**

**Ques 1(b). 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:- **Given

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 = 120 × 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 contains 10Ω. What resistance must be added in series for the circuit to have a current of 5.0 A?**

**Answer:- **Given

Supply Voltage V = 120 V

Resistance = 10 ohm

Require current = 5 A

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

**V = IR**

120 = 5 × (10 + R)

24 = 10 + R

R = 24 − 10

**R = 14 ohm**

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

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

**Answer:- ** Given

R_{1} = 9Ω, R_{2} = 4Ω

R_{4} = 12Ω , R_{5} = 36Ω

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)Ω**

Resistance **R _{4}** and

**R**are in parallel hence equivalent resistance

_{5}**R _{eq2} = (R_{4} × R_{5})/(R_{4} + R_{5})**

= (12 × 36)/(12 + 36) = 432/48 = **9Ω**

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 = **12Ω—————(2)**

Equating equation 1 and 2 we get

\begin{array}{l}\dfrac{{(9 + {R_3})4}}{{13 + {R_3}}} + 9 = 12\\\\\dfrac{{36 + 4{R_3}}}{{13 + {R_3}}} = 3\\\\36 + 4{R_3} = 39 + 3{R_3}\\\\{R_3} = 3\Omega \end{array}

**⇒ 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}

\begin{array}{l} = \dfrac{{It \times ({R_1} + {R_3})}}{{({R_1} + {R_3}) + {R_2}}}\\\\ = \dfrac{{1 \times (9 + 3)}}{{(9 + 3) + 4}}\\\\{I_2} = 0.75A\end{array}

Hence the voltage across resistance R2

**V _{2} = I_{2}(R_{2}) = 0.75 × 9 = 3V**

or

**V _{2} = V_{equ} = R_{3} × I’_{equ} = 3 × 1 = 3V**

**⇒ Calculation of Current I _{4}**

Since Resistance R4 and R5 in parallel

So using current division rule

I_{4} = I_{R4}

\begin{array}{l}{I_4} = \dfrac{{{I_t} \times {R_5}}}{{{R_4} + {R_5}}}\\\\ = \dfrac{{1 \times 36}}{{12 + 36}} = \dfrac{{36}}{{48}}\\\\ = 0.75A\end{array}

So R_{3} = 3Ω

V_{2} = 3V

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

or

V_{4} = V_{t} − V_{2}

= 12 − 3 = 9V

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

Therefore

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