NEET Class neet Answered
solve the problem
![question image](http://images.topperlearning.com/topper/new-ate/top_mob1655377621250745961a28b3b43-b148-4c18-9b35-bdeddd34e427.jpg)
Asked by pankajpankaj55224 | 16 Jun, 2022, 16:37: PM
Power dissipated in resistor = i2 R
where i is the current passing through resistor of resistance R
if 36 W is dissipated in 9 Ω resistor , then current i through resistor is calculated as follows
i2 × 9 = 36
hence we get current i =
A = 2 A
![begin mathsize 14px style square root of 36 divided by 9 end root end style](https://images.topperlearning.com/topper/tinymce/cache/119e546436efcea4ed24ce11a49e3898.png)
Voltage across 9 Ω resistor = i × R = 2 × 9 = 18 V
Since 6 Ω resistor is in parallel to 9 Ω resistor , voltage across 6 Ω resistor also 18 V
Hence current through 6 Ω resistor = voltage / resistance = (18 / 6 ) A = 3 A
Hence current drawn from battery = ( 3 + 2 ) A = 5 A
Voltage across 2 Ω resistor = i × R = 5 × 2 = 10 V
Answered by Thiyagarajan K | 16 Jun, 2022, 20:36: PM
NEET neet - Physics
Asked by shaikmuneerjaan | 18 Jul, 2024, 20:58: PM
NEET neet - Physics
Asked by anshgupta8840400152 | 01 Jun, 2024, 23:36: PM
NEET neet - Physics
Asked by hardikmittal25 | 03 May, 2024, 14:57: PM
NEET neet - Physics
Asked by sojusvi | 17 Apr, 2024, 13:12: PM
NEET neet - Physics
Asked by mohitsisodia612 | 19 Mar, 2024, 21:21: PM
NEET neet - Physics
Asked by puchkiii431 | 16 Mar, 2024, 21:43: PM
NEET neet - Physics
Asked by riyachabhadiya5 | 03 Jan, 2024, 14:49: PM
NEET neet - Physics
Asked by ameetapradhan55 | 10 Dec, 2023, 20:39: PM
NEET neet - Physics
Asked by dhivi242004 | 02 Apr, 2023, 11:04: AM
NEET neet - Physics
Asked by parvathalupeddapuli | 24 Dec, 2022, 20:25: PM