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Current Electricity question

2019 · 12 Jan · Shift 1 · Q68

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Current Electricity
2019 · 12 Jan · Shift 1 · Q68

Current Electricity question

2019 · 12 Jan · Shift 1 · Q68

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1

Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then :

A
P1 = 4W, P2 = 16 W
B
P1 = 16W, P2 = 4 W
C
P1 = 9W, P2 = 16 W
D
P1 = 16W, P2 = 9 W

View written solutionFree

Correct answer: B

Find the resistances of the bulbs from their ratings

For a bulb rated $(P, V)$ , its resistance is

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

So,

For the $25\,\text{W}, 220\,\text{V}$ bulb:

R_1 = \frac{220^2}{25} = \frac{48400}{25} = 1936\,\Omega

For the $100\,\text{W}, 220\,\text{V}$ bulb:

R_2 = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega

Since the bulbs are in series, the same current flows through both

Total resistance:

R_{\text{total}} = R_1 + R_2 = 1936 + 484 = 2420\,\Omega

Current in the series circuit:

I = \frac{V}{R_{\text{total}}} = \frac{220}{2420} = \frac{1}{11}\,\text{A}

Calculate power consumed by each bulb

Using

P = I^2 R

Power in the $25\,\text{W}$ bulb:

P_1 = I^2 R_1 = \left(\frac{1}{11}\right)^2 \cdot 1936

P_1 = \frac{1936}{121} = 16\,\text{W}

Power in the $100\,\text{W}$ bulb:

P_2 = I^2 R_2 = \left(\frac{1}{11}\right)^2 \cdot 484

P_2 = \frac{484}{121} = 4\,\text{W}

Match with the options

We get:

P_1 = 16\,\text{W}, \quad P_2 = 4\,\text{W}

So the correct option is B.

Check with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.

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