Question III.B.2: (a) Find the value of R and the current through each branch ...
(a) Find the value of R and the current through each branch in the circuit shown in Fig. 39 if the current in branch AO is zero.
(b) Calculate the current flowing through the 6 Ω resistor in the circuit shown in Fig. 40
(a) The circuit is redrawn as shown in Fig. 42. Since this is a bridge circuit and current through x is zero, the bridge is balanced.
\frac{P}{Q}=\frac{R}{S}Hence,
or \frac{1}{1.5}=\frac{4}{R}
or R= 6Ω
Current through x is zero. Therefore, A and O are at the same potential. These two points can be put together as shown in Fig. 4
Current, I=\frac{10}{0.8+1.2+2}= \frac{10}{4}=2.5 A
This current is being supplied by the battery. The current through branch MON is I_{1} and through branch MAN is I_{2}. These two branch currents are:
I_{1}=2.5× \frac{1}{4+1}= \frac{2.5}{5}=0.5 A
I_{2}=2.5× \frac{4}{4+1}= \frac{10}{5}=2.5 A
(b) Applying KVL in loop ABEFA
10-2I_{1}-6(I_{1}-I_{2})-6-4I_{1}=0
or 6I_{1}+3I_{2} =2 (i)
Applying KVL in loop BCDEB
-3I_{2}-2-5I_{2}+6+6(I_{2}-I_{1})=0
or 3_{1}+I_{2}=-1 (ii)
From Eqs. (i) and (ii)
6 I_{1}+3I_{2}=26 I_{1}+2I_{2}=-2
Subtracting I_{2} =4A
From Eq. (ii)
3I_{1}=-I_{2}-1=-4-1=-5or I_{1} =-\frac{5}{3} A
