Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

​four bulbs B 1, B 2 ,B 3 ,B 4of 100w each are connected to220v main.the reading in an ideal ammeter will be

​four bulbs B1,B2,B3,B4of 100w each are connected to220v main.the reading in an ideal ammeter will be

Grade:12

2 Answers

Arun
25763 Points
one year ago
We know that P = V2/R Here V across all the bulbs is 220 V and P = 100 W. Then the resistance of the bulb is R = V2/P = (220)2/100 = 48400/100 = 484 Ω The circuit when reduce gives 4 resistors each of resistance R = 484 Ω in parallel to each other. Hence, the equivalent resistance is R/4 = 484/4 Ω = 121 Ω Total current in the circuit is V/R = 220/121 = 20/11 and this is equally divided among the bulbs. Current flowing into B2, B3 and B4 will be reading of the ammeter = 20/11 x 3/4 = 15/11 A = 1.36 A
Vikas TU
14149 Points
one year ago
Dear student 
Current through each bulb  =100/220=5/11A
Therefore current through ideal ammeter  = 3×5/11=1.35A    
Good Luck 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free