Question

a 100W,120V lamp is connected in series with another lamp of 40W,120V and the combination is connected across mains.Calculate the value of the resistance to be connected across the second lamp,so that each lamp may get the proper current at the rated voltage.

Answer 1

Answer:

The resistance that is connected across second bulb will be 540 ohm

Explanation:

As we know that the resistance of two bulbs is given as

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

$R_1 = \frac{120^2}{100}$

$R_1 = 144 ohm$

and we have

$R_2 = \frac{120^2}{40}$

$R_2 = 360 ohm$

Now the maximum current through each bulb is given as

$P = VI$

$100 = 120 I_1$

$I_1 = 0.83 A$

for another bulb we have

$40 = 120 I_2$

$I_2 = 0.33 A$

now when two bulbs are connected in series such that a resistance R is in parallel with second bulb then we have

$R = R_1 + R_2$

$R = 144 + \frac{360R}{360 + R}$

$\frac{120}{0.33} = 144 + \frac{360R}{360 + R}$

$216 = \frac{360R}{360 + R}$

$360 + R = 1.67 R$

$R = 540 ohm$

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Topic : Power Rating of bulb

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