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?

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/Pn = 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?

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 R3, V2, and I4.

Vt = 12 V, It = 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

Req1 = R1 + R3 = (9 + R3)Î©

Resistance R4 and R5 are inÂ  parallel hence equivalent resistance

Req2 = (R4Â Ã— R5)/(R4 + R5)

= (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 Req = (Req1 || 4) + Req2

Req = [(9 + R3) || 4] + 9 ——————(1)

Also Rt = Vt/It = 12/1 = 12Î©—————(2)

Equating equation 1 and 2 we get

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

â‡’ Voltage across Resistance R2

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

I2 = IR2

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

Hence the voltage across resistance R2

V2 = I2(R2) = 0.75Â Ã— 9 = 3V

or

V2 = Vequ = R3Â Ã— I’equ = 3Â Ã— 1 = 3V

â‡’ Calculation of Current I4

Since Resistance R4 and R5 in parallel

So using current division rule

I4 = IR4

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

V2 = 3V

I4 = 0.75 A

or

V4 = VtÂ âˆ’ V2

= 12Â âˆ’ 3 = 9V

I4 = V4/R4 = 9/12 = 0.75 A

Therefore

R3 = 3Î©, V2 = 3V, I4 = 0.75A

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