The maximum permissible current through a galvanometer G of internal resistance \(10\ \Omega\) is 0.05A. A resistance R is used to convert G into a voltmete...
Question 1 Report
The maximum permissible current through a galvanometer G of internal resistance \(10\ \Omega\) is 0.05A. A resistance R is used to convert G into a voltmeter with a maximum reading of 100V. Find the value of R and how it is connected to G
Answer Details
The question is asking for the value of the resistor R that needs to be connected to a galvanometer G of internal resistance 10 ohms, to convert it into a voltmeter with a maximum reading of 100V. The maximum current that can flow through the galvanometer is given as 0.05A.
To convert the galvanometer into a voltmeter, a resistor needs to be connected in series with it. The value of this resistor can be calculated using the formula:
R = (Vmax/Ig) - Rg
Where R is the value of the resistor to be connected, Vmax is the maximum voltage that the voltmeter can read (in this case, 100V), Ig is the maximum current that can flow through the galvanometer (0.05A), and Rg is the internal resistance of the galvanometer (10 ohms).
Substituting the given values, we get:
R = (100/0.05) - 10
R = 1990 ohms
Therefore, the value of the resistor R that needs to be connected in series with the galvanometer G is 1990 ohms. It should be connected in series with the galvanometer to convert it into a voltmeter.
So, the correct answer is "1990 ohms in series".
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning