# SSC JE Electrical Conventional Paper Solved 2015-Electrical-Exam

# SSC JEÂ Electrical Conventional Paper with Explained SolutionÂ 2015 | MES Electrical

**Ques 1(a).Â **A copper wire has a resistance of 0.85Î© at 20Â°C. What will be its resistance at 40Â°C? The temperature coefficient of resistance of copper at 0Â°C is 0.004Â°C.

**Answer.Â ** Resistance Temperature Coefficient

**R _{t}Â =Â R_{o}(1 +Â Î±_{o}t)**

Where

R_{o}Â is the resistance at 0Â° Celsius

R_{t}Â is the Resistance at tÂ° Celsius.

Given R_{t} =Â 0.85Î© at 20Â°C

Temperature coefficient of resistanceÂ Î± = 0.04Â°C

tÂ = 20Â°

R_{o} = ?

0.85 =Â R_{o} [ 1 + 0.004 Ã— 20]

0.85 =Â R_{o}Ã—1.08

**R _{o} = 0.787Â Î©**

Now the resistance of the copper at 40Â°C

**R _{t}Â =Â R_{o}(1 +Â Î±_{o}t)**

Given R_{t} =Â ? at 40Â°C

Temperature coefficient of resistanceÂ Î± = 0.04Â°C

tÂ = 20Â°

R_{o} = 0.787 Î©

R_{t} =Â 0.787 [ 1 + 0.004 Ã— 40]

**[R _{t} = 0.912 Î©]**

**Answer**

**Ques 1(b).Â **In the circuit shown in figure what is the value of V_{B}?

**Answer:-** In the above figure the two voltage sources are connected in series therefore there equivalentÂ voltage will be (V_{1} + V_{2})

Then the total voltage in the given question will be

[V_{1} + (âˆ’V_{2})] = (6Â âˆ’ 5) = 1V

V_{eq} = IR

Or IR = 1

Now potential difference across the resistorÂ

V_{A}Â âˆ’ V_{B} = IR

6Â âˆ’ V_{B} = 1

**âˆ´ V _{B} = 5V**

**Ques 1(c).Â **What is the value of Thevenin voltage E_{TH} in the given circuit as shown in the figure?

**Answer:–Â ** As per Thevenin theorem, when resistance R_{L} is connected across terminals A and B, the network behaves as a source of voltage E_{Th} and internal resistance R_{T} and this is called Thevenin equivalent circuit.

**Thevenin Voltage**

The Thevenin voltage e used in Thevenin’s Theorem is an ideal voltage source equal to the open circuit voltage at the terminals. In the given question, the resistance 1Î© does not affect this voltage and the resistances 3Î© and 7Î©Â form a voltage divider, giving

**Ques 1(d):-Â **In figure find the value of resistance R

**Answer:**

Applying KCL at Node “1”

2V_{1}Â âˆ’ 100 + 20 = 0

2V_{1} = 80

V_{1} = 40 V

R = V_{1}/I = 40/2 = 20Î©