wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The current I through a rod of a certain metallic oxide is given by I=0.2 V5/2, where V is the potential difference across it. The rod is connected in series with a resistance to a 6V battery of negligible internal resistance. What value (in ohms) should the series resistance have so that the current in the circuit is 0.44? (round off the value to the nearest integer)

Open in App
Solution

Voltage drop across the RodI=0.2(V)52

0.44=0.2(V)52

2.2=(V)52

V=(2.225)

logV=25×log(2.2)

logV=0.4×0.34=0.14VV=1.15volt

Now voltage at point X61.15=4.85volt

apply V=IR on the resistance R(4.850)=0.44RR=11.02Ω

Nearest intrger value=11Ω

329131_126038_ans.jpg

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Alternating Current Applied to a Capacitor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon