Test: Kirchhoff’s Current & Voltage Law

The current through the 10 ohm resistor = v₁ / 10 = 2A

Applying KCL at node 1: i₅ = i₁₀ + i₂

i₂ = 6 - 2 = 4A

Thus the drop in the 2 ohm resistor = 4×2 = 8V

v₁ = 20V; hence v₂ = 20-v across 2 ohm resistor = 20 - 8 = 12V
v₂ = v since they are connected in parallel.
v = 12V

KCL states that the total current leaving the junction is equal to the current entering it. In this case, the current entering the junction is 5 A + 10 A = 15A

Assume lower terminal of 20 ohm at 0V and upper terminal at V volt and applying KCL, we get V/10 + V/20 = 1

V = (20 / 3) V

So, current through 20 ohm = V/20 = (20/3) / 20 = 1/3 = 0.33V

 According to KCL, I₁ + I₂ + I₃ = 0

Hence I₃ = - (I₁ + I₂) = - 5A

•  At junction a: i₁ - i₃ - i₂ = 0
  i₂ = 2A
• At junction b: i₄ + i₂ - i₆ = 0
  i₄ = -1A
• At junction c: i₃ - i₅ + i₄ = 0
  i₅ = 2A

Current = Charge/Time

Here charge = 50C and time = 5 seconds

so current = 50/5 = 10A

KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge.

KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes.

KCL is applied for different nodes of a network whether it is planar or non-planar.

 At the junction, I - 2 + 3 - 4 - 5 = 0

Hence I = 8A

Using KVL, 12 - V1 - 8 = 0.

V1 = 4V

8 - V2 - 2 = 0

V2 = 6V

KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge.

Using voltage divider rule, V = 10*12/30 = 4V

Using KVL, the matrix to find the loop currents are:
MATRIX 10 -100 * i1 + 100 * i2 = 0
MATRIX  10 - 200 * i2 + 100 * i1 = 0
I1 = 0.3, I2 = 0.2

According to KVL, the sum of the voltage over any closed loop is equal to 0.

Kirchhoff’s laws, namely Kirchhoff’s Current Law and Kirchhoff’s Voltage law are the basic laws in order to analyze a circuit.

According to Kirchhoff’s Voltage Law, Every mesh is a loop but every loop is not a mesh.

V_eq = 10 + 5 -20 = -5u
R_eq = 5 + 2 + 3 = 10Ω
I = V/R = -5/10 = -0.5A.

For branch A: V_AC= 15 * 20 / (25+15) = 7.5V
For branch B: V_BC= 10 * 20 / (10+40) = 4V
Applying KVL to loop ABC
V_AB+V_BC+V_CA = 0
V_AB = 3.5V

Mesh analysis helps us to utilize the different voltages in the circuit as well as the I_R products in the circuit which is nothing but KVL.