Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - Filling the Blanks (Q.No. 3)
3.
In the given circuit, the current flowing through R3 equals ______.
Discussion:
3 comments Page 1 of 1.
Sapana M A said:
8 years ago
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.
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:
8 years ago
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.
-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.
Deepak jr said:
9 years ago
How this? Explain me.
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