What is the resistivity (in ohms-m) of a 2 ohm cylindrical wire when the length and the diameter of the wire are 10 m and 0.4 m respectively?

The resistance of the conductor is determined by the 
R = ρL/A
Where
ρ = Resistivity of the conductor
L = Length of the conductor = 10 m
A = Area of the conductor = πR2 = 3.14 × 0.22 = 0.125 m2
R = Resistance = 2 Ω
Therefore the resistivity is
ρ = A × R ⁄ L
= 0.125 × 2 ⁄ 10
ρ = 0.025 Ω-m

 What will be the equivalent resistance (in Ω) for the circuit given below?

In the given circuit the resistance R9 and R8 are Parallel with Resistance R10 therefore
(R9 + R8) || R10
= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}
= (6 × 12) ⁄ (6 + 12)
RA = 4Ω
Now resistance RA and R7 is parallel with Resistance R6
∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}
= (12 × 6) ⁄ (12 + 6)
RB = 4Ω
Now the circuit becomes as shown in the figure
Resistance RB and R7 is parallel with Resistance R4
{(4 + 8) × 6} ⁄ {(4 + 8) + 6}
= (12 × 6) ⁄ (12 + 6)
RC = 4Ω
Now resistance RC and R3 is parallel with Resistance R2
{(4 + 2) × 6} ⁄ {(4 + 2) + 6}
RD = 3
Now our final circuit becomes as shown in the figure
Therefore the equivalent resistance is
R1 + RD
Req = 4 + 3 = 7Ω
 

 Which of the following statement is CORRECT?

Although the Thevenin’s theorem and Norton’s theorem can be used to solve a given network, yet the circuit approach differs in the following respects:
A Norton’s theorem is converse (opposite) of Thevenin’s theorem in the respect that Norton equivalent circuit uses a current generator instead of the voltage generator and the resistance RN (which is the same as RTH) in parallel with the generator instead of being in series with it.
Thevenin’s theorem is a voltage form of an equivalent circuit whereas Norton’s theorem is a current form of an equivalent circuit.
To Convert Thevenin equivalent circuit into Norton’s equivalent circuit the following step is involved
RN = RTH
IN = ETH ⁄ RTH
 

What is the value of current (in A) for the given junction?
 

According to Kirchhoff’s Current Law: At any point in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.
 
From the above Diagram 
Current Flowing towards the Point: I2, I6, I4 
Current Flowing Away from the Point: I1, I3, I5
Hence I2 + I6 + I4 = I1 + I3+ I5
Putting the value of the current
2A + 7A + I4 = 4A + 3A + 8A
I4 = 15A − 9A = 6A
I4 = 6A

 Two wires of the same resistivity have equal length. The cross-sectional area of first wire is two times to the area of the other. What will be the resistance (in ohms) of the wire that has a large cross-sectional area, if the resistance of the other wire is 20 Ohms?

What will be the value of Thevenin’s voltage (in V). The Thevenin’s resistance (in Ω) and the load current (in A) respectively, across the load resistor in the given electrical circuit?

As per Thevenin theorem, when resistance RL is connected across terminals A and B, the network behaves as a source of voltage ETh and internal resistance RT and this is called Thevenin equivalent circuit.
Thevenin Voltage
The Thevenin voltage 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 10Ω does not affect this voltage and the resistances 30Ω and 20Ω form a voltage divider, giving
 
\begin{array}{l}{E_{Th}} = 100 \times \dfrac{30}{{30 + 20}}\\\\{E_{TH}} = 60V\end{array}
Thevenin’s resistance can be found by replacing 100 V source with a short-circuit.
Thevenin equivalent resistance for the given network is
R = (R1 || R2) + R3
Rth = (20 || 30) + 10 = (20 x 30) ⁄ (20 + 30) + 10 = 22Ω
Rth = 22Ω
The Load current Is calculated as
IL = ETH ⁄ (RTH + RL)
= 60 ⁄ (22 + 5) = 2.22 A
Hence the value of Thevenin voltage, Thevenin Resistance, and Load current is (60 V, 22Ω, 2.22A) respectively.