Question

Given: I_A = 2 mA, I_D = - 0.1 mA, I_G = I_H = I_K

(a) Find the numerical value of I_B in mA.

(b) Find the numerical value of I_H in mA.

(c) If V_A = - 3 V and V_G = 2 V, find the numerical value of the power in element B in mW and whether it is supplying or receiving power.

(d) If also V_F = - 100 mV, what is the power in element H in mW?

(e) If also VE = - 150 mV, what is the power in element C and D together in mW?

Answer 1

(a) Element A and B lie in same branch of the circuit. Therefore, the magnitude of the current is same in both elements. The direction of current in the elements is opposite. Therefore, 13 =-11 =-(2 mA) =-2 mA Element C and D lie in same branch of the circuit. Therefore, the magnitude of the current is same in both elements. Therefore, Ic = ID =-0.1 mA Use junction rule at point, where element B, C,G,H, and K are connected as follows: 16 +1, +Ix = 10-13 Since, 16 =1, EIK 31,=1c-1 =-0.1 mA-(-2 mA) Therefore 1, -1.9 mA =10.63 mA Consider the loop containing elements A, B, and G. Apply Kirchhoff's loop law, while considering the signs of the voltage shown figure.

-V+V-VG=0 V, =Va+V = 2 V+3 V =5 V Therefore, the magnitude of V, is 5 V. The current through the element is -2 mA. Therefore, power through the element B is given as follows: |PBI=11:13 =(sv)(-2 mA)( 1 ) = 10-4 W Since, current is entering through the positive terminal, the element is absorbing or receiving the power. (d) Consider the loop containing elements G, H, and F. Apply Kirchhoff's loop law, while considering the signs of the voltage shown figure. VG-V,+V;=0 V=VG+V; =2 V +(100 mV) Topy = 2.1 V Therefore, the magnitude of V, is 2.1 V. The current through the element is 0.63 mA. Therefore, power through the element B is given as follows:

|PH=\V#1: =(2,1V)(0.63 mA)(1 = 13 W