Question

Problem 4. Use Kirchhoffs laws and Ohm's law to calculate the current i2 flowing across the 6k2 resistor and V2in Figure 3. Show all the work. 54k1V 2 kC2 30i 8V Figure 3

Answer 1

Apply KCL at node 1: - V_2 - 6 V - 1 V/54 K + V_2/6K + V_2 - 8 V - V/2K = 0 -(1) LCM of 54 K, 6K, 2 K is 54 K therefore V_2 - 7V + 9 V_2 + 27 V_2 - 27 times 8 - 27 V/54 K = 0 37 V_2 - 223 V - 27 V = 0 -(2) V = 30 i_1 (From the above Fig) i_1 = 7 V - V_2/54 K v = 30 (7V - V_2)/ 54 K = 5 (7 V - V_2)/9K -(3) Substitute eq (3) in eq (2), it becomes 37 V_2 - 223 V - 27 times 5 (7 V - V_2)/9K = 0 \r\n 37 V_2 - 223 V = 15 (7 V - V_2)/1000 37000 V_2 - 223000 V = 105 V - 15 V-2 37015 V_2 = 223105 V V_2 = 223105/37015 V V_2 = 6.0274 V i_1 = V_2/6K i_2 = 6.0274/6K i_2 = 0.001 A i_2 almostequalto 1 mA