(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 : Calculate : (i) the value of current through each resistor. (ii) the total current in the circuit. (iii) the total effective resistance of the circuit.

Foundation

Physics Foundation

Ohms Law and Resistance

Question

(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 :
Calculate :
(i) the value of current through each resistor.
(ii) the total current in the circuit.
(iii) the total effective resistance of the circuit.

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(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 the current flowing through the whole circuit is I.
By ohm's law,

\frac{V}{I} = R

V = I × R ............(2)
Applying ohm's law to both R₁, R₂ and R₃
V = I × R₁ ............(3)
V = I × R₂ ............(4)
V = 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 5Ω = R₁, 10Ω = R₂, 30Ω = R₃
(i) Current through

R_{1} = I_{1} = \frac{V}{R_{1}} = \frac{6}{5} = 1.2 A

Current through

R_{2} = I_{2} = \frac{V}{R_{2}} = \frac{6}{10} = 0.6 A

Current through

R_{3} = I_{3} = \frac{V}{R_{3}} = \frac{6}{30} = 0.2 A

(ii) Total current in the circuit = 1.2 + 0.6 + 0.2 = 2 A
(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 Ω

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