CBSE Class 10 Science 2019 · Paper 31/1/1 · Question 20(a)

Electricity 5 marks Medium

(a)With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
(b)In an electric circuit two resistors of $12 Ω$ each are joined in parallel to a 6 V battery. Find the current drawn from the battery.

Answer

The same $V$ across each resistor and $I = I_{1} + I_{2} + I_{3}$ give $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ ; two $12 Ω$ resistors in parallel draw 1 A.

(a) Circuit: resistors

R_{1}

R_{2}

and

R_{3}

are connected between the same two points X and Y; a battery, a plug key K and an ammeter A are in the main line, and a voltmeter V is connected across X and Y. The potential difference

V

across each resistor is the same, and the main current divides:

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

. By Ohm's law

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

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

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

, and for the equivalent resistance

R_{p}

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

. So

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

, and dividing by

V

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

. (b)

\frac{1}{R _{p}} = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}

, so

R_{p} = 6 Ω

. Current drawn from the battery:

I = \frac{V}{R _{p}} = \frac{6 V}{6 Ω} = 1

🔒 Step-by-step solution

CBSE Class 10 Science 2019 · Paper 31/1/1 · Question 20(a)

Answer

From the board papers

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