Help me find current in AB branch using nodal analysis (kcl).
Answers
Answer:
Solution
Given :-
\sf \implies \: Equation \: 3x + 4y - 9 = 0⟹Equation3x+4y−9=0
\sf \implies Point \: (1,3) \: and \: (2,7)⟹Point(1,3)and(2,7)
Let
\sf \implies \: Ratio \: = P \ratio \: 1⟹Ratio=P:1
Using section formula
\implies \sf \: p \bigg( \dfrac{x_2m + nx_1}{m + n} , \: \dfrac{y_2m + y_1n}{m + n} \bigg)⟹p(
m+n
x
2
m+nx
1
,
m+n
y
2
m+y
1
n
)
Where
\begin{gathered} \sf \implies \: x_1 = 1, y_1 = 3 \\ \sf \implies \: x_2 = 2,y_2 = 7\end{gathered}
⟹x
1
=1,y
1
=3
⟹x
2
=2,y
2
=7
\sf \implies \: m = p \: \: and \: n = 1⟹m=pandn=1
Now put the value on formula
\sf \implies\: p \bigg( \dfrac{2 \times p + 1 \times1}{p + 1} , \: \dfrac{7 \times p + 3 \times 1}{p + 1} \bigg)⟹p(
p+1
2×p+1×1
,
p+1
7×p+3×1
)
\sf \implies\: p \bigg( \dfrac{2p + 1 }{p + 1} , \: \dfrac{7 p + 3 }{p + 1} \bigg)⟹p(
p+1
2p+1
,
p+1
7p+3
)
Now put the value of x and y on Given equation
\sf \implies \: 3x + 4y - 9 = 0⟹3x+4y−9=0
\sf \implies3 \bigg( \dfrac{2p + 1}{p + 1} \bigg) + 4 \bigg( \dfrac{7p + 3}{p + 1} \bigg) - 9 = 0⟹3(
p+1
2p+1
)+4(
p+1
7p+3
)−9=0
\sf\implies \: \dfrac{6p + 3}{p + 1} + \dfrac{28p + 12}{p + 1} - 9 = 0⟹
p+1
6p+3
+
p+1
28p+12
−9=0
Takin Lcm
\sf \implies \: \dfrac{6p + 3 + 28p + 12 - 9p - 9}{p + 1} = 0⟹
p+1
6p+3+28p+12−9p−9
=0
\sf \implies\: 6p + 3 + 28p + 12 - 9p - 9 = 0⟹6p+3+28p+12−9p−9=0
\sf \implies \: 34p + 15 - 9p - 9 = 0⟹34p+15−9p−9=0
\sf \implies 25p + 6 = 0⟹25p+6=0
\sf \implies \: p = \dfrac{ - 6}{25}⟹p=
25
−6
Answer
The ratio is -6/25 or -6:25