Question

Determine the total current I in the Fig. 2?

: 50

Marks

10

Solve the given network circuit by using Mesh analysis to find V0 and Vx and also find the voltage across each resistor element in Fig.1?

2 

Answer 1

Using the method of Super mesh analysis.

Considering loop 1,

$12k\Omega \ I_2 + 4k\Omega \ (I_2 - I_1) + 8k\Omega \ (I_2 - I_3) = 0 \\ 12k\Omega \ I_2 + 4k\Omega \ I_2 - 4k\Omega \ I_1 + 8k\Omega \ (I_2 - I_3) = 0 \\ 24k\Omega \ I_2 - 4k\Omega \ I_1 - 8k\Omega \ I_3 = 0 -----eqn(1)\\$

$Since \ I_3 - I_1 = 6mA --------- eqn(2) \\ \therefore I_3 = 6mA + I_1 \\ substituting \ for \ I_3 \ in \ equation (1) \\ 24k\Omega \ I_2 - 4k\Omega \ I_1 - 8k\Omega \ (6mA + I_1) = 0 \\ 24k\Omega \ I_2 - 4k\Omega \ I_1 - 8 k\Omega \ I_1 - 48 = 0 \\ 24k\Omega \ I_2 - 12k\Omega \ I_1 - 48 = 0 ------ eqn (3) \\$

Considering loop 2,

$-6V_x + 4k\Omega (I_1 - I_2) + 8k\Omega (I_3 - I_2) + 12k\Omega \ I_3 \\ since, \ V_x = 8k\Omega (I_3 - I_2) -------eqn (4) \\ -48k\Omega (I_3 - I_2) + 4k\Omega I_1 - 12k\Omega I_2 + 20k\Omega I_3 = 0 \\ 4k\Omega \ I_1 - 36k\Omega \ I_2 - 28k\Omega \ I_3 = 0 ---- eqn (5) \\ substituting \ for \ I_3 \ in \ equation(5) \\ 4k\Omega \ I_1 - 36k\Omega \ I_2 - 28k\Omega(6mA + I_1) = 0 \\ 4k\Omega \ I_1 - 64k\Omega \ I_2 - 168 = 0 ----- eqn (6) \\$

Solving equation (3) and equation (6) using simultaneously;

$-12k\Omega \ I_1 + 24k\Omega \ I_2 = 48 ------- eqn (3) \\ 4k\Omega \ I_1 - 64k\Omega \ I_2 = 168 ------ eqn (6) \\ I_1 = -0.0106 \ A \\ I_2 = -3.2857 \times 10^{-3} \ A = -3.2857 \ mA \\ Since \ I_3 - I_1 = 6mA ------ \ from \ eqn(2) \\ I_3 = 6mA + I_1 \\ I_3 = 6 \times 10^{-3} \ A - 0.0106 \ A \\ I_3 = -4.6 \times 10^{-3} A = -4.6 \ mA \\ \therefore \\ (i) \\ V_o = I_3 \times 12k\Omega \\ V_o = - 4.6 \times 10^{-3} \ A \ \times 12k\Omega \\ V_o = 55.2 \ V \\ (ii) \\ V_x = 8k\Omega (I_3 - I_2) \\ V_x = 8k\Omega (-4.6 \ mA - (-3.2857 \ mA)) \\ V_x = 8\times 10^{3} \times -1.143 \times 10^{-3} \\ V_x = 10.5144 \ V \\ (iii) Voltage \ across \ resistor \ 4k\Omega; \\ V = 4k\Omega (I_1 - I_2) V = 29.26 V \\ (iv) \\ Voltage \ across \ resistor \ 12k\Omega; \\ V = 12k\Omega \times I_2 \\ V = 39.43 \ V$