CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Figure shows two resistors R1 and R2 connected to a battery having an emf of 40V and negligible internal resistance. A voltmeter having a resistance of 300Ω is used to measure potential difference across R1. Find the reading of the voltmeter?
675331_49868c832a5749199935cf24e1a66c35.png

Open in App
Solution

as given in the circuit, current is given by I
I=VReq
Req=R2+R1RvR1+Rv where Rv is the resistance of the given voltmeter
Req=880+200×300200+300=1000Ω
I=VReq=401000=0.04volt
Reading of volmeter = potential difference =
I×(R1×RvR1+Rv)I×200×300200+300)=0.04×120=4.8volt

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conductivity and Resistivity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon