# SSC JE Electrical Conventional Paper with Explained Solution 2015 | MES Electrical

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

**Answer. ** Resistance Temperature Coefficient

**R _{t} = R_{o}(1 + α_{o}t)**

Where

R_{o} is the resistance at 0° Celsius

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

Given R_{t} = 0.85Ω at 20°C

Temperature coefficient of resistance α = 0.04°C

t = 20°

R_{o} = ?

0.85 = R_{o} [ 1 + 0.004 × 20]

0.85 = R_{o}×1.08

**R _{o} = 0.787 Ω**

Now the resistance of the copper at 40°C

**R _{t} = R_{o}(1 + α_{o}t)**

Given R_{t} = ? at 40°C

Temperature coefficient of resistance α = 0.04°C

t = 20°

R_{o} = 0.787 Ω

R_{t} = 0.787 [ 1 + 0.004 × 40]

**[R _{t} = 0.912 Ω]**

**Answer**

**Ques 1(b). **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 equivalent voltage will be (V_{1} + V_{2})

Then the total voltage in the given question will be

[V_{1} + (−V_{2})] = (6 − 5) = 1V

V_{eq} = IR

Or IR = 1

Now potential difference across the resistor

V_{A} − V_{B} = IR

6 − V_{B} = 1

**∴ V _{B} = 5V**

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

**Answer:– ** 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 1Ω does not affect this voltage and the resistances 3Ω and 7Ω form a voltage divider, giving

\begin{array}{l}{E_{Th}} = 50 \times \dfrac{7}{{7 + 3}}\\\\{E_{TH}} = 35V\end{array}

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

**Answer:**

Applying KCL at Node “1”

\dfrac{{{V_1} - 100}}{{10}} + \dfrac{{{V_1}}}{{10}} + 2 = 0

2V_{1} − 100 + 20 = 0

2V_{1} = 80

V_{1} = 40 V

R = V_{1}/I = 40/2 = 20Ω