STATE & EXPLANATION (KCL)
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STATE & EXPLANATION (KVL)
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EXAMPLE (KCL)
Find i and v in the given circuit
Applying KCl at node a, and, v = iR
0.5i + 3 = i => v = 6 * 4
=> 0.5i = 3 => v = 24 V
=> i = 6 A Ans :
Ans :
4Ω
EXAMPLE (KVL)
Find i and v
x
in the given circuit
Here , V
x
= 10i,
Applying KVL, -35 + 10i + 2V
x
+ 5i = 0
=> -35 + 10i + 20i+ 5 i= 0
=> 35i = 35
=> i = 1 A
Ans:
Then, V
x
= 10i
=> V
x
= 1v
Ans:
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i
APPLICATION (KCL)
KCL
ÞTo combine current sources in parallel
Total Current, I
T
+ I
2
= I
1
+ I
3
=> I
T
= I
1
- I
2
+ I
3
(B)
APPLICATION (KVL)
KVL
ÞTo find the total voltage
Total voltage,
-V
ab
+ V
1
- V
2
= 0
=> V
ab
= V
1
- V
2
(A)
(B)
V
ab