CBSE Class 10 Science 2020 · Paper 31/C/1 · Question 29(b)

Electricity 5 marks Medium

(a)Derive the relation for the equivalent resistance when three resistors of resistances $R_{1}$ , $R_{2}$ and $R_{3}$ are connected in parallel.
(b)Find the minimum resistance that can be made using four resistors, each of $20 Ω$ .

Answer

For parallel resistors $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ ; four $20 Ω$ resistors in parallel give the minimum, $5 Ω$ .

(a) Let

R_{1}

R_{2}

and

R_{3}

be connected in parallel between points X and Y, with a battery maintaining a potential difference

V

across X and Y. In parallel, the potential difference across each resistor is the same,

V

, and the total current

I

divides among them:

I = I_{1} + I_{2} + I_{3}

. By Ohm's law,

I_{1} = \frac{V}{R _{1}}

I_{2} = \frac{V}{R _{2}}

and

I_{3} = \frac{V}{R _{3}}

, and for the equivalent resistance

R_{p}

I = \frac{V}{R _{p}}

. Hence

\frac{V}{R _{p}} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}}

, which gives

\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

. (b) The resistance is least when all four resistors are connected in parallel:

\frac{1}{R} = \frac{4}{20} = \frac{1}{5}

, so the minimum resistance is

5 Ω

🔒 Step-by-step solution

CBSE Class 10 Science 2020 · Paper 31/C/1 · Question 29(b)

Answer

From the board papers

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