(a) Find the potential difference across AB and across AC?
(b) Find the current through the other two resistors?
(c) Find the total resistance?
The given network, shown in the figure a, can be represented as shown in figure b.
The resultant of 4 Ω and 8 Ω resistances, which are in series in the arm EDC, is given by
R1 = 4 Ω + 8 Ω = 12 Ω
Similarly, the resultant of 4 Ω and 8 Ω in the arm EFC is given by
R2 = 4 Ω + 8 Ω = 12 Ω
The equivalent circuit of R1 and R2 which are in parallel is shown in figure c.
If R is the equivalent resistance between the point A and B, then
Or R = 6 Ω
Potential difference across AB = V1 = I1R1 = 1 × 5 = 5V
Since R2 and R3 are in parallel, resultant resistance between B and C is given by
Potential difference across BC, i.e.,
V2 = I1 RP = 1 × 6 = 6V (∵ current through BC is I1, i.e., 1 A)
Potential difference across AC
= Potential difference across AB + Potential difference across BC = 5 V + 6 V = 11 V
(b) current through R2, i.e.,
(∵ Potential difference across R2 is the same as across BC)
Current through , i.e., I3 = I1 – I2 = 1A – 0.6 = 0.4 A
(c) Total resistance between A and C, i.e., R = R1 + RP = 5Ω + 6Ω = 11Ω
The resultant of R2 and R3 which are in parallel is RP; and R1 and RP are in series
R1 = 5Ω + 20Ω = 25Ω
R2 is the equivalent resistance of 15Ω and 10Ω which are in series,
R2 = 15 Ω + 10 Ω = 25 Ω
Now R1 and R2 are in parallel so final resistance is given by
Total potential difference across the arrangement of all the resistances is 24 V
Total current in the circuit,
Since RP and 6Ω are in series, total resistance of the combination, i.e.,
R = 6 + RP = 6 + 1.2 = 7.2Ω
Rs = 3Ω + 3Ω + 3Ω = 9Ω
Further, Rs (= 9Ω) and R5 (= 3Ω) are in parallel so their equivalent resistance is given by
Thus, the equivalent resistance between C and D is 2.25 Ω.
(ii) Now R1 (= 3Ω), RP (= 2.25 Ω)
and R6 (= 3 Ω)
Are in series
Thus, equivalent resistance between A and B,
= R1 + RP + R6
= 3 + RP + R6
= 3 + 2.25 + 3
= 8.25 Ω