Electricity Worksheet-7
(a) Find the potential difference across AB and across AC?
(b) Find the current through the other two resistors?
(c) Find the total resistance?
Answer:
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
R_{1} = 4 Ω + 8 Ω = 12 Ω
Similarly, the resultant of 4 Ω and 8 Ω in the arm EFC is given by
R_{2} = 4 Ω + 8 Ω = 12 Ω
The equivalent circuit of R_{1} and R_{2} 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 = V_{1} = I_{1}R_{1} = 1 × 5 = 5V
Since R_{2} and R_{3} are in parallel, resultant resistance between B and C is given by
Potential difference across BC, i.e.,
V_{2} = I_{1} R_{P} = 1 × 6 = 6V (∵ current through BC is I_{1}, 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 R_{2}, i.e.,
(∵ Potential difference across R_{2} is the same as across BC)
Current through , i.e., I_{3} = I_{1} – I_{2} = 1A – 0.6 = 0.4 A
(c) Total resistance between A and C, i.e., R = R_{1} + R_{P} = 5Ω + 6Ω = 11Ω
The resultant of R_{2} and R_{3} which are in parallel is R_{P}; and R_{1} and R_{P }are in series
R_{1} = 5Ω + 20Ω = 25Ω
R_{2} is the equivalent resistance of 15Ω and 10Ω which are in series,
R_{2} = 15 Ω + 10 Ω = 25 Ω
Now R_{1} and R_{2} 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,
,
i.e.,
Since R_{P} and 6Ω are in series, total resistance of the combination, i.e.,
R = 6 + R_{P} = 6 + 1.2 = 7.2Ω
R_{s} = 3Ω + 3Ω + 3Ω = 9Ω
Further, R_{s} (= 9Ω) and R_{5} (= 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 R_{1} (= 3Ω), R_{P} (= 2.25 Ω)
and R_{6} (= 3 Ω)
Are in series
Thus, equivalent resistance between A and B,
= R_{1} + R_{P} + R_{6}
= 3 + R_{P} + R_{6}
= 3 + 2.25 + 3
= 8.25 Ω