Question

Draw the circuit diagram for an electric bulb, rheostat and

a combination of cells​

Answer 1

Please refer the attachment for the simple circuit which consist of :

1. Electric bulb

2. Ammeter

3.Plug key

4. Cell or battery ( group of cells)

Current flows when there is a potential difference between 2 ends in a circuit.

Always current flows from positive terminal to negative terminal.

Ammeter should be connected in series in circuit.

More information:-

Resistors in series :

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Parallel connection :

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Closed circuit :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8,1){\line(1,0){0.5}}\put(10.4,1){\line(1,0){0.6}}\put(8.7,1){\line(1,0){1.2}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){1.3}}\put(9.7,3){\line(3,0){1.3}}\put(8.5,0.7){\line(0,1){0.5}}\put(8.7,0.8){\line(0,1){0.3}}\put(9.7,1){\sf{\large{ \bigg(}}}\put(10.2,1){\sf{\large{ \bigg)}}}\put(9.5,3){\circle{0.4}}\put(9.4,2.9){\line(3,2){0.3}}\put(9.6,2.8){\line( - 2,3){0.2}}\put(8.1,1.1){\sf{ + }}\put(8.7,1.1){\sf{ \large{ - }}}\put(10.18,1){\circle*{0.1}}\put(8.5,0.2){\sf{ \large{Closed circuit }}}\end{picture}$

Open Circuit :

$\setlength{\unitlength}{1.5cm}\begin{picture}(0,0)\linethickness{0.4mm}\put(8,1){\line(1,0){0.5}}\put(10.4,1){\line(1,0){0.6}}\put(8.7,1){\line(1,0){1.2}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){1.3}}\put(9.7,3){\line(3,0){1.3}}\put(8.5,0.7){\line(0,1){0.5}}\put(8.7,0.8){\line(0,1){0.3}}\put(9.7,1){\sf{\large{ \bigg(}}}\put(10.2,1){\sf{\large{ \bigg)}}}\put(9.5,3){\circle{0.4}}\put(9.4,2.9){\line(3,2){0.3}}\put(9.6,2.8){\line( - 2,3){0.2}}\put(8.1,1.1){\sf{ + }}\put(8.7,1.1){\sf{ \large{ - }}}\put(8.5,0.2){\sf{ \large{O{p}en circuit }}}\end{picture}$

Answer 2

Explanation:

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