Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Using Kirchoff's rules, calculate the current through the $40\; \Omega$ and $20\; \Omega$ resistors in the following circuit.

Answers (1)

According to Kirchoff's loop rule, The sum of potential differences across a closed loop is zero.

In ABCDA, applying Kirchoff's Loop Rule -

$80-20I_{1}-40I_{2}=0$

$8-2I_{1}-4I_{2}=0$

On rearranging, the equation we have,

$I_{1}+2I_{2}-4=0$ ___(1)

In $CDEFC,$ applying Kirchoff's Loop Rule -

$-40I_{2}-40+10 (I_{1}-I_{2})=0$

$-40\; I_{2}-40+10\; I_{1}-10 \; I_{2}=0$

$-50\; I_{2}-40+10\; I_{1}=0$

On rearranging, the equation we have

$I_{1}-5\; I_{2}-4=0$ _____(2)

from (1) and (2)

$I_{1}-5\; I_{2}-4-I_{1}+2I_{2}-4=0$

$7\; I_{2}=0$

$I_{2}=0$

On equaling the value of $I_{2}$ in equation (1) , we get

$I_{1}=4A$

Hence, there is no current through the $40\Omega$ resistor and $4A$ current flows through the $20\Omega$ resistor.

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th

anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question