CBSE Class 10 Science 2019 · Paper 31/2/1 · Question 20(b)

Electricity 5 marks Medium

(a)Three resistors $R_{1}$ , $R_{2}$ and $R_{3}$ are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.
(b)Calculate the equivalent resistance of the following network :

Answer

For resistors in parallel $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ ; the equivalent resistance of the network is $20 Ω$ .

(a) Circuit diagram:

R_{1}

R_{2}

and

R_{3}

are joined between two common points X and Y; the battery, the plug key and the ammeter are in series in the main line, and the voltmeter is connected across X and Y (see the figure). Expression: the total current divides among the branches,

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

. The potential difference

V

is the same across each resistor, so by Ohm's law

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

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

and

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

. If

R_{p}

is the equivalent resistance,

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

. Hence

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

, that is,

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

. (b) The two

20 Ω

resistors in parallel:

\frac{1}{R _{p}} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10}

, so

R_{p} = 10 Ω

. This is in series with

10 Ω

: equivalent resistance of the network

= 10 Ω + 10 Ω = 20 Ω

🔒 Step-by-step solution

CBSE Class 10 Science 2019 · Paper 31/2/1 · Question 20(b)

Answer

From the board papers

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