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
347 views
in Physics by (93.7k points)
closed by
In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is `80 ohm`. The reading of the voltmeter will be
image
A. 0.80 volt
B. 1.60 volt
C. 1.33 volt
D. 2.00 volt

1 Answer

0 votes
by (94.0k points)
selected by
 
Best answer
Correct Answer - C
(c ) Total resistance of the circuit `= (80)/(2) + 20 = 60 Omega`
`implies` Main current `i= (2)/(60) = (1)/(30)`
Combination of voltmeter and `80 Omega` resistance is connected in series with `20 Omega`, so current through `20 Omega` and this combination will be same `= (1)/(30) A`.
Since the resistance of voltmeter is also `80 Omega`, so this current is equally distributed in `80 Omega` resistance and voltmeter (i.e., `1//60 A` through each).
`P.D.` across `80 Omega` ressitance `= (1)/(60) xx 80 = 1.33 V`

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

...