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
165 views
in Physics by (90.7k points)
closed by
A `0.2 k Omega` resistor and `15 mu F` capacitor are connected in series to a 220 V, 50 Hz ac source. The impadance of the circuit is
A. `250 Omega`
B. `268 Omega`
C. `29.15 Omega`
D. `291.5 Omega`

1 Answer

0 votes
by (91.8k points)
selected by
 
Best answer
Correct Answer - D
Here, `R=0.2 k Omega = 200 Omega`
`C =15 mu F = 15xx10^(-6)F, V_("rms")=220 V, upsilon = 50 Hz`
Capacitive reactance,
`X_(C )=(1)/(2pi upsilon C)=(1)/(2xx3.14xx50xx15xx10^(-6))=212 Omega`
The impedance of the Rccircuit is
`Z = sqrt(R^(2)+X_(C )^(2))=sqrt((200)^(2)+(212)^(2))=291.5 Omega`

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

...