We will find the value of the current I as read by the ammeter first by applying series parallel circuits technique and then by using mesh resistance matrix.
1. Series Parallel Circuit Technique
The total impedance as seen by the voltage source is
= 1 + [j1 || (2 - j1)] = 1 + (j1(2 - j1))/2 = 15 + j1
∴ total circuit current i = (5 ∠0º)/(1.5 + j1)
This current gets divided into two parts at point A, one part going through the ammeter and the other going along AB. By using current-divider rule. we have
I = 5/(1..5 + j1) x J1/(2 + j1 - j1) = j5/(3 + j2)
2. Mesh Resistance Matrix
In Fig. 15.48 (b),
R11 = (1 + j1), R22 = (2 + j1 −j1) = 2; R12 = R21 = −j1
As shown in Fig., the voltage source has been interchanged with the ammeter. The polarity of the voltage source should be noted in particular. It looks as if the voltage source has been pushed along the wire in the counterclockwise direction to its new position, thus giving the voltage polarity as shown in the figure. We will find the value of I in the new position of the ammeter by using the same two techniques as.
1. Series Parallel Circuit Technique
As seen by the voltage source from its new position, the total circuit impedance is
= 2(2 - j1) + j1 || 1 = (3 + j2)/(1 + j1)
The total circuit current i = 5 x (1 + j1)/(3 + j2)
This current i gets divided into two parts at point B as per the current-divider rule.
I = (5(1 + j1))/(3 + j2) x j(1/(1 + j1)) = j5/(3 + j2)
2. Mesh Resistance Matrix
As seen from Fig. .
