Derive the formula for resultant resistance when three resistors are connected in
A) series
B) parallel
Answers
Answer:
three resistances R1, R2 and R3 Connected in series with a battery of let the p.d. across R1, R2 and R3 is V1, V2 and V3 respectively.
S.t. V = V1 + V3 + V2 -------------(1)
Let the equivalent resistance be R and current flowing through whole circuit is 1.
By ohm’s law,
V/1
V = I × R -----------(2)
Applying ohm’s law to both R1, R2 and R3,
V1 = I × R1 --------(3)
V2 = I × R2 ---------(4)
V3 = I × R3 ----------(5)
From eqs. (1), (2), (3), (4) and (5), we get
I × R = I × R1 + I × R2 + I × R3
I × R = I × (R1 + R2 + R3)
R = R1 + R2 + R3
(b) Let 5 Ω = R1, 10 Ω = R2, 30 Ω = R3
(i) Current through R1 = I1 = V/R1 = 6/5 = 1.2A
Current through R2 = I2 = V/R2 = 6/10 = 0.6 A
Current through R3 = I3 = V/R3 = 6/30 = 0. 2 A
(ii) Total current in the circuit = 1.2 + 0.6 + 0.2 = 2A
(iii) Effective resistance R is given as
1/R = 1/R1 + 1/R2 + 1/R3
= 1/5 + 1/10 + 1/30
= (6 + 3 + 1)/30 = 10/30
R = 30/10 = 3 Ω