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## Class 10 Physics (India)

### Course: Class 10 Physics (India)>Unit 3

Lesson 8: Electric circuit with Bulbs

# Solved example: Power dissipated in bulbs

Let's tackle questions in which we have circuits with devices (like bulbs) whose power ratings are mentioned.  Created by Mahesh Shenoy.

## Want to join the conversation?

• Can I use P=VI for the given rating to find the current flowing in the bulb and then use that current to find the dissipated power?
• Actually, we first need to find the resistance of each bulb, then find the equivalent resistance of all the resistors, THEN we find the current flowing through the entire circuit.
Because, sometimes the resistances can be in a parallel resistor setup, and then the current of the circuit will differ from the current of the each resistor.
Hope this helps. :)
• Why the bulb one which had a lower power rating was predicted to be glowing brighter than the one which had a higher power rating?
(1 vote)
• Both the bulbs get the same amount of current flow. Then consider the following cases :

CASE 1 BULB WITH HIGHER POWER - Needs higher current flow to glow at full brightness

CASE 2 BULB WITH LOW POWER - Since the power required is less even a lower current will just be enough to make it glow brighter.

This is my logic. I hope someone points me out if am wrong.
• Is the Ali's question now wrong as the correct answer is coming 9.6 W but it's showing 9.8 W ?
• Yes, the answer given by Khan Academy is wrong. The answer should be 9.6 W. So you're correct.
(1 vote)
• Shouldn't we also add the fact in the video that because the power dissipated in bulb2 is way higher than rated wattage, it is likely to burn out? And hence, the right statement would be that bulb2 burns brighter momentarily and then fuses off and no current flows throught the circuit thereafter - and neither bulb glows?
(1 vote)
• Why did the bulb with the lower power rating glows brighter?
(1 vote)
• The main reason is that..

The one with a LOWER power rating requires less energy as compared to the bulb with a HIGHER power rating (Which requires higher power to glow as bright.)

But since the current is equal the one with the lesser power rating will glow brighter.
(1 vote)
• In Q2, if we go 1 step ahead and try to find the voltage across bulb 2, we will find the voltage is = V = IR = 4 A x 8 Ohm = 32 V. Now as per bulb rating the maximum voltage it can have is 20 V but in this circuit it is going till 32 V, which means it will break at this voltage. So with a battery of 40V, this circuit will break as bulb 2 will not be able to hold 32v of voltage. Please evaluate this and advice if this is the correct observation.
(1 vote)
• What is the voltage across each bulb
(1 vote)
• how to calculate heat in resistor
• `H=VIt`
`H=V^2t/R [I=V/R]`
`H=I^2Rt [V=IR]`