Question

For the figure shown below find the current and voltage across all resistors and the cell (use Kirchhoff's laws)

Answer 1

In the figure, on applying Junction Law we get,

$\sf{\longrightarrow I_1=I_2+I_3\quad\quad\dots(1)}$

On applying Loop Law in left loop,

$\sf{\longrightarrow 4I_3+3I_1=12}$

From (1),

$\sf{\longrightarrow 4I_3+3(I_2+I_3)=12}$

$\sf{\longrightarrow 7I_3+3I_2=12\quad\quad\dots(2)}$

On applying Loop Law in right loop,

$\sf{\longrightarrow 4I_3-2I_2=5\quad\quad\dots(3)}$

Multiplying (2) by 2,

$\sf{\longrightarrow 14I_3+6I_2=24\quad\quad\dots(4)}$

Multiplying (3) by 3,

$\sf{\longrightarrow 12I_3-6I_2=15\quad\quad\dots(5)}$

Adding (4) and (5),

$\sf{\longrightarrow26I_3=39}$

$\sf{\longrightarrow\underline{\underline{I_3=\dfrac{3}{2}\ A}}}$

From (2),

$\sf{\longrightarrow I_2=\dfrac{12-7I_3}{3}}$

$\sf{\longrightarrow I_2=\dfrac{12-\dfrac{21}{2}}{3}}$

$\sf{\longrightarrow\underline{\underline{I_2=\dfrac{1}{2}\ A}}}$

From (1),

$\sf{\longrightarrow I_1=\dfrac{1}{2}\ A+\dfrac{3}{2}\ A}$

$\sf{\longrightarrow\underline{\underline{I_1=2\ A}}}$

Voltage across $\sf{4\ \Omega}$ resistor $\sf{=4I_3}=\sf{\underline{\underline{6\ V}}.}$

Voltage across $\sf{3\ \Omega}$ resistor $\sf{=3I_1}=\sf{\underline{\underline{6\ V}}.}$

Voltage across $\sf{2\ \Omega}$ resistor $\sf{=2I_2}=\sf{\underline{\underline{1\ V}}.}$