AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two bulbs A and B are rated $100 W - 120 V$ and $10 W - 120 V$ respectively. They are connected in parallel across a $120 V$ source. Which bulb will consume more energy?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Bulb A

Bulb B

Both 1 and 2

None of the above

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The current through each bulb as these are in series. Bulb B will consume more energy.
Given
Bulb A is rated as

= 100 W - 120 V

Bulbs B is rated as

= 10 W - 120 V

Two bulbs are connected across a source in a series of

= 120 V

In bulb A - Voltage

= 120 V

Power

= 100 W

In bulb B-
Voltage

= 120 V

Power

= 10 W

Now let us find the resistance of the two bulbs -

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

Let us find the resistance for the bulb A-
Bulb A (R) =

\frac{12 0^{2}}{100}

= 144 ohm

Bulb B (r) =

\frac{12 0^{2}}{10}

= 1440 ohm

Now we got the resistance for both the bulbs.
To find the series combination, we need to use the formula -

R = R₁ + R₂

\Rightarrow R = 144 + 1440

\Rightarrow R = 1584 ohm

is the resistance in the series combination.
When these bulbs in series are connected across

120 V

, current in the circuit, i.e.,

I = \frac{V}{R}

= \frac{120}{1584}

= 0.076 A

So 10 watts bulb consumes more energy because uses less power and also the amount of
current flowing is less.
Remember the formula-

P = I \times E

.
Power consumed by bulb A,
i.e.,

P_{1} = I^{2} \times R_{1}

{\Rightarrow P}_{1} = (0.076 A)^{2} \times (144 Ω)

{\Rightarrow P}_{1} = 0.832 W

Power consumed by bulb B,
i.e.,

P_{2} = I^{2} {\times R}_{2}

{\Rightarrow P}_{2}

= (0.076 A)^{2} \times (1440 Ω)

{\Rightarrow P}_{2}

= 8.32 W

Hence, the bulb B (rated

10 W - 120 V

) consumes more energy than the bulb A (rated

100 W - 120 W

) when they are connected in series.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!