Q.9.    A battery of 9 V is connected in series with resistors of 0.2\Omega , 0.3\Omega ,0.4\Omega , 0.5\Omega and 12\Omega  respectively. How much current would flow through the 12\Omega  resistor?                 
 

Answers (1)

Total resistance  =R

R=0.2+0.3+0.4+0.5+12=13.4\Omega

V = 9V

I=\frac{V}{R}=\frac{9}{13.4}=0.67 A

Hence, All resistors are in series so 0.67A current would flow through the 12\Omega  resistor.                 

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions