Test: Basic Laws- 2

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

Nodes = 10 (a, b, c, d, e, f, g, h, k, p)
Junction points = 3(b, e, h)

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