Electronics - Series-Parallel Circuits - Discussion

3. 

In the given circuit, the current flowing through R3 equals ______.

[A]. 10 mA
[B]. 21.11 mA
[C]. 90 mA
[D]. 100 mA

Answer: Option A

Explanation:

No answer description available for this question.

Deepak Jr said: (Oct 4, 2016)  
How this? Explain me.

Sapana M A said: (Sep 16, 2017)  
By applying KVL for both loops we get following equations.

200I1-100I2=19,
-100I1+ 1000I2= -19,
so we get I1=90mA.
and I2=-10mA,
So across R3 current I2 is flowing which is 10mA.

Sapana Adhalli said: (Sep 20, 2017)  
200 I1 - 100 I2= 19 --> equation 1.
-100 I1 + 1000 I2 = -19 --> equation 2.
solving equation 1 and 2 we get,
I1 =90mA and I2 which is accross R3 that is I2 = -10mA.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.