Three resistors $R_{1}, R_{2}, R_{3}$ are joined in series. Draw labelled diagram and obtain expression of its equivalent resistance.

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Solution

In Series combination of resistances, the second end to each resistance is connected to the first end of the next resistance and so on.
In a series combination of resistances, the current flowing through each resistance will be the same and will be equal to the current supplied by the battery. As resistance is different and the current flowing through each resistance is the same, hence potential differences across different resistances will be different.
Let three resistance $R_{1}, R_{2}, R_{3}$ are connected in series with a cell of potential difference $V$ . Let $I$ current supplied by the battery in this combination.
Let,
$V_{1} =$ Potential difference across the ends of resistance $R_{1}$
$V_{2} =$ Potential difference across the ends of resistance $R_{2}$ .
$V_{3} =$ Potential difference across the ends of resistance $R_{3}$ .
$V = V_{1} + V_{2} + V_{3}$ ........(1)
If equivalent resistance of the combination be $R$ then by Ohm's law,
$V = I R, V_{1} = I R_{1}, V_{2} = I R_{2}, V_{3} = I R_{3}$
Put in Eq.(1)
$I = I R_{1} + I R_{2} + I R_{3}$
$I R = I (R_{1} + R_{2} + R_{3})$
$R = R_{1} + R_{2} + R_{3}$
Therefore, in a series combination of resistance equivalent resistance is equal to the sum of all individual resistances.

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Write the expressions for the equivalent resistance R of three resistors $R_{1}, R_{2}$ and $R_{3}$ joined in (a) parallel, and (b) series.

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