Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.9k views
in Physics by (101k points)
closed by
A battery of emf 2.5 V and internal resistance r is connected is series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA. Draw the circuit diagram and calculate the value of r.

1 Answer

0 votes
by (100k points)
selected by
 
Best answer
Circuit diagram for the given data is shown below.
image
Given, `E = 2.5 V, R = 4 5 Omega`,
`r_(A) = 1A, I = 50 mA`
r = ?
`E = I (R + r_(A) + r)`
`2.5 = 50 xx 10^(-3) (45 + 1 + r)`
`46 + r = (2.5)/(50 xx 10^(-3)) = (2.5 xx 10^(3))/(50) = 50`
`:. r = 50 - 46 = 4 Omega`

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...