Test: Basic Laws- 2 - Electrical Engineering (EE) MCQ

Let the current flowing through 5 Ω resistor be I' Applying KCL, we get
5 + I = I'.
Applying KVL, we get

So, 
= -4.4 A

Explanation:

Superposition Theorem

• It states that in any linear, active, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance.
• The superposition theorem is used to solve the network where two or more sources are present and connected.
• It can be applied to Linear bilateral networks.
• The bilateral circuit allows the current flow in both directions.
• The transmission line is the main example of the bilateral circuit because if you give a power supply from any direction, the circuit properties remain constant

Thevenin Theorem

• It can be applied in complex or large networks to find current and voltage in few branches.
• Thevenin’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.

Kirchhoff's voltage and current laws

These are used to determine currents and voltage in all branches of a network.

The maximum power transfer theorem

• It helps in Impedance matching in audio circuits.
• The Maximum Power Transfer Theorem is another useful circuit analysis method to ensure that the maximum amount of power will be dissipated in the load resistance when the value of the load resistance is exactly equal to the resistance of the power source. 

• The maximum power transfer can be obtained even if the output impedance is not the same as the load impedance.
• This can be done using a suitable “turns ratio” on the transformer with the corresponding ratio of load impedance.

Applying KCL at the node connecting 0.5 V current source and 1 A current source, we get
1 + I = 0.5 V
For v - 2 volt, 1 + I = 1
or, I = 0 A

Using KVL in the loop, we have
V₁ = 10 + 15 = 25 volts
Also, V₂ = -25 volts
∴ V₁ + V₂ = 25 - 25 = 0 volts

Let the required voltage be V
Then, voltage across 10 Ω resistor

or,
Hence, required voltage is
V = 180 volts

Using Ohm’s law,

Given, V₁ = 2V₂
As I is same therefore,
R₁ = 2R₂
∴ R_T = R₁ + R₂ = 2R₂ + R₂ = 3R₂
So, R_T = 3R₂ = 4Ω
or, 
and 

We know that
R_Δ = 3 R_Y
or, 

On combining the current sources, the circuit is reduced as shown below.

Hence, voltage drop across the 2Ω resistor
= 2 x 2 = 4 volts

Using KCL,
I₂ = 6 - 2 = 4 A
Also, 
or, 
and I₄ = I₂ - I₃ = 4 - (-1.5) = 5.5 A
Also, I₁ = 6 A
∴ I₁ + I₂ + I₃ + I₄ = 6 + 4 - 1.5 + 5.5
= 15.5 - 1.5 = 14A 

= (10 + j10 + j20)
= (10 + j30) Ω

Given, I_25Ω=4 A
Here, R' = 80║50║R



or, 
or, 

or, 29R + 400 = 40R or 11R = 400
or, R = 36.36 Ω

Here, a and c are at equipotential.
Also, b and d are at equipotential
∴ 

Number of different node-pair voltage = Number of KCL equations = n - 1 =10 - 1 = 9