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

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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.

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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.

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Ques 1(d). In the following circuit, find the total resistance R₃, V₂, and I₄.

Answer:-  Given

R₁ = 9Ω, R₂ = 4Ω
R₄ = 12Ω , R₅ = 36Ω

V_t = 12 V, I_t = 1 A

R₃ = ?, V₂ = ?, I₄ = ?

The given circuit can be reconstructed as shown in the figure

The Resistance R₁ and R₃ are in series therefore equivalent Resistance is

R_eq1 = R₁ + R₃ = (9 + R3)Ω

Resistance R₄ and R₅ are in  parallel hence equivalent resistance

R_eq2 = (R₄ × R₅)/(R₄ + R₅)

= (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

[latex]\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}[/latex]

⇒ Voltage across Resistance R₂

Now resistance (R₁ + R₃) and R₂ are in parallel therefore by using current divison rule

I₂ = I_R2

[latex]\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}[/latex]

Hence the voltage across resistance R2

V₂ = I₂(R₂) = 0.75 × 9 = 3V

or

V₂ = V_equ = R₃ × I’_equ = 3 × 1 = 3V

⇒ Calculation of Current I₄

Since Resistance R4 and R5 in parallel

So using current division rule

I₄ = I_R4

[latex]\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}[/latex]

So R₃ = 3Ω
V₂ = 3V

I₄ = 0.75 A

or

V₄ = V_t − V₂

= 12 − 3 = 9V

I₄ = V₄/R₄ = 9/12 = 0.75 A

Therefore

R₃ = 3Ω, V₂ = 3V, I₄ = 0.75A