# Two identical resistors each of **resistance 10 ohm **are connected in:

**(i) Series**, **(ii) Parallel**, in turn to a battery of** 6V**.

Calculate the **ratio of power** consumed by the combination of resistor in the two cases.

30656"

Given:

Two resistances of 10 ohm each are connected in Series and then Parallel.

Voltage = 6 V.

To find: The ratio of power consumed in the two cases

Solution:

We know power consumed by a resistor $ = \frac{Voltage^2}{Resistance}$

Or

$P = \frac{V^2}{R}$

Series case:

Effective Resistance R = R_{1} + R_{2}

= 10 + 10

So effective resistance R = 20 ohm

Power $P = \frac{V^2}{R}$

$= \frac{6^2}{20} = 1.8$

Power = 1.8 watt

Parallel case:

We can find effective resistance using the formula

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{10} + \frac{1}{10}$

$\frac{1}{R} = \frac{2}{10}$

So effective resistance, $R = \frac{10}{2}$

= 5 ohm

Power $P = \frac{V^2}{R}$

$= \frac{6^2}{5} = 7.2$

Power = 7.2 watt

Ratio of power in the given cases = $\frac{1.8\ watt}{ 7.2 \ watt}$

= $\frac{1}{4}$

So ratio of power consumed is $\frac{1}{4}$

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