(a) With the help of a diagram, derive the formula for the resultant resistance of three resistors connected in series.
(b) For the circuit shown in the diagram given below :
(a) Fig shows three resistances R1, R2 and R3 connected in series with a battery of V volts.
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 I.
By Ohm's law,
\(\frac{V}{I}\) = R
V = I x R ........(2)
Applying Ohm's law to both R1, R2 and R3
V1 = I x R1 ........(3)
V2 = I x R2 ........(4)
V3 = I x R3 .........(5)
From eqs. (1), (2), (3), (4) and (5), we get
I x R = I x R1 + I x R2 + I x R3
I x R = I x (R1 + R2 + R3)
R = R1 + R2 + R3
(b) Let 5 ohms = R1, 10 ohm = R2, 30 ohms = R3
(i) Current through R1 = I2 = \(\frac{V}{R_1}\)
= \(\frac{6}{5}\) = 1.2 A
Current through R2 = I2 = \(\frac{V}{R_2}\)
= \(\frac{6}{10}\) = 0.6 A
Current through R3 = I3 = \(\frac{V}{R_3}\)
= \(\frac{6}{30}\) = 0.2 A
(ii) Total current in the circuit = 1.2 + 0.6 + 0.2 = 2A
(iii) Effective resistance R is given as
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)
= \(\frac{1}{5}\) + \(\frac{1}{10}\) + \(\frac{1}{30}\)
= \(\frac{6 + 3 + 1}{30}\)
= \(\frac{10}{30}\)
R = \(\frac{30}{10}\) = 3 ohm