Question

Find current through each branch
[Hint : Use Kirchhoff Voltage Law and Kirchhoff Junction Law]


Pls help me...


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Answer 1

Let,

• net current in the circuit $\sf{=i.}$

• current through AB $\sf{=i_{AB}.}$

• current through AD $\sf{=i_{AD}.}$

• current through BC $\sf{=i_{BC}.}$

• current through DC $\sf{=i_{DC}.}$

Applying Junction Law at A,

$\longrightarrow\sf{i=i_{AB}+i_{AD}\quad\quad\dots(1)}$

Applying Junction Law at B,

$\longrightarrow\sf{i_{AB}=i_{BC}+i_{BD}\quad\qud\dots(2)}$

Applying Junction Law at C,

$\longrightarrow\sf{i_{BC}+i_{DC}=i\quad\quad\dots(3)}$

Applying Junction Law at D,

$\longrightarrow\sf{i_{AD}+i_{BD}=i_{DC}\quad\quad\dots(4)}$

Applying Voltage Law in the loop EADCFE,

$\longrightarrow\sf{10=10\,i+5\,i_{AD}+10\,i_{DC}\quad\quad\dots(5)}$

Applying Voltage Law in the loop EADBCFE,

$\longrightarrow\sf{10=10\,i+5\,i_{AD}-5\,i_{BD}+5\,i_{BC}\quad\quad\dots(6)}$

Applying Voltage Law in the loop EABDCFE,

$\longrightarrow\sf{10=10\,i+10\,i_{AB}+5\,i_{BD}+10\,i_{DC}\quad\quad\dots(7)}$

Applying Voltage Law in the loop EABCFE,

$\longrightarrow\sf{10=10\,i+10\,i_{AB}+5\,i_{BC}\quad\quad\dots(8)}$

Substituting (1) and (3) in (5),

$\longrightarrow\sf{10=10\,i+5\left(i-i_{AB}\right)+10\left(i-i_{BC}\right)}$

$\longrightarrow\sf{10=25\,i-5\,i_{AB}-10\,i_{BC}}$

$\longrightarrow\sf{2=5\,i-\left(\,i_{AB}+2\,i_{BC}\right)}$

$\longrightarrow\sf{\,i_{AB}+2\,i_{BC}=5\,i-2\quad\quad\dots(9)}$

From (8),

$\longrightarrow\sf{2\,i_{AB}+i_{BC}=2-2\,i\quad\quad\dots(10)}$

Solving (9) and (10) we get,

$\longrightarrow\sf{i_{BC}=4i-2\quad\quad\dots(11)}$

From (10),

$\longrightarrow\sf{i_{AB}=2-3i\quad\quad\dots(12)}$

From (1),

$\longrightarrow\sf{i_{AD}=4i-2\quad\quad\dots(13)}$

From (3),

$\longrightarrow\sf{i_{DC}=2-3i\quad\quad\dots(14)}$

From (8),

$\longrightarrow\sf{10=10\,i+10(2-3i)+5(4i-2)+10(2-3i)}$

$\longrightarrow\sf{10=30-30i}$

$\longrightarrow\underline{\underline{\sf{i=\dfrac{2}{3}\ A}}}$

From (11) and (13),

$\longrightarrow\underline{\underline{\sf{i_{AD}=i_{BC}=\dfrac{2}{3}\ A}}}$

From (12) and (14),

$\longrightarrow\underline{\underline{\sf{i_{AB}=i_{DC}=0\ A}}}$

Answer 2

Answer:

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