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?
2Â
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 \\\\\n12k\\Omega \\ I_2 + 4k\\Omega \\ I_2 - 4k\\Omega \\ I_1 + 8k\\Omega \\ (I_2 - I_3) = 0 \\\\\n24k\\Omega \\ I_2 - 4k\\Omega \\ I_1 - 8k\\Omega \\ I_3 = 0 -----eqn(1)\\\\"
"Since \\ I_3 - I_1 = 6mA --------- eqn(2) \\\\\n\\therefore I_3 = 6mA + I_1 \\\\\nsubstituting \\ for \\ I_3 \\ in \\ equation (1) \\\\\n24k\\Omega \\ I_2 - 4k\\Omega \\ I_1 - 8k\\Omega \\ (6mA + I_1) = 0 \\\\\n24k\\Omega \\ I_2 - 4k\\Omega \\ I_1 - 8 k\\Omega \\ I_1 - 48 = 0 \\\\\n24k\\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 \\\\\nsince, \\ V_x = 8k\\Omega (I_3 - I_2) -------eqn (4) \\\\\n-48k\\Omega (I_3 - I_2) + 4k\\Omega I_1 - 12k\\Omega I_2 + 20k\\Omega I_3 = 0 \\\\\n4k\\Omega \\ I_1 - 36k\\Omega \\ I_2 - 28k\\Omega \\ I_3 = 0 ---- eqn (5) \\\\\nsubstituting \\ for \\ I_3 \\ in \\ equation(5) \\\\\n4k\\Omega \\ I_1 - 36k\\Omega \\ I_2 - 28k\\Omega(6mA + I_1) = 0 \\\\\n4k\\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) \\\\\n4k\\Omega \\ I_1 - 64k\\Omega \\ I_2 = 168 ------ eqn (6) \\\\\nI_1 = -0.0106 \\ A \\\\\nI_2 = -3.2857 \\times 10^{-3} \\ A = -3.2857 \\ mA \\\\\nSince \\ I_3 - I_1 = 6mA ------ \\ from \\ eqn(2) \\\\ \nI_3 = 6mA + I_1 \\\\\nI_3 = 6 \\times 10^{-3} \\ A - 0.0106 \\ A \\\\\nI_3 = -4.6 \\times 10^{-3} A = -4.6 \\ mA \\\\\n\\therefore \\\\\n(i) \\\\ \nV_o = I_3 \\times 12k\\Omega \\\\\nV_o = - 4.6 \\times 10^{-3} \\ A \\ \\times 12k\\Omega \\\\\nV_o = 55.2 \\ V \\\\\n(ii) \\\\\nV_x = 8k\\Omega (I_3 - I_2) \\\\\nV_x = 8k\\Omega (-4.6 \\ mA - (-3.2857 \\ mA)) \\\\\nV_x = 8\\times 10^{3} \\times -1.143 \\times 10^{-3} \\\\\nV_x = 10.5144 \\ V \\\\\n(iii) Voltage \\ across \\ resistor \\ 4k\\Omega; \\\\\nV = 4k\\Omega (I_1 - I_2)\nV = 29.26 V \\\\\n(iv) \\\\\nVoltage \\ across \\ resistor \\ 12k\\Omega; \\\\ \nV = 12k\\Omega \\times I_2 \\\\\nV = 39.43 \\ V"
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