Question

Using kirchoffs laws find the currents and voltages across all the resistors in the figure shown below

Answer 1

Applying Junction Law we get,

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

Applying Loop Law in left loop,

$\sf{\longrightarrow 4I_1+2I_2=-28}$

From (1),

$\sf{\longrightarrow 4(I_2-I_3)+2I_2=-28}$

$\sf{\longrightarrow4I_3-6I_2=28\quad\quad\dots(2)}$

Applying Loop Law in right loop,

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

Multiplying (3) by 3,

$\sf{\longrightarrow 3I_3+6I_2=-21\quad\quad\dots(4)}$

Adding (2) and (4),

$\sf{\longrightarrow 7I_3=7}$

$\sf{\longrightarrow\underline{\underline{I_3=1\ A}}}$

From (3),

$\sf{\longrightarrow I_2=-\dfrac{I_3+7}{2}}$

$\sf{\longrightarrow\underline{\underline{I_2=-4\ A}}}$

From (1),

$\sf{\longrightarrow I_1=I_2-I_3}$

$\sf{\longrightarrow\underline{\underline{I_1=-5\ A}}}$

[Negative signs in $\sf{I_1}$ and $\sf{I_2}$ show that they're in opposite direction to that depicted in the figure.]

Voltage across $\sf{R_1,}$

$\sf{\longrightarrow V_1=I_1R_1}$

$\sf{\longrightarrow\underline{\underline{V_1=-20\ V}}}$

Voltage across $\sf{R_2,}$

$\sf{\longrightarrow V_2=I_2R_2}$

$\sf{\longrightarrow\underline{\underline{V_2=-8\ V}}}$

Voltage across $\sf{R_3,}$

$\sf{\longrightarrow V_3=I_3R_3}$

$\sf{\longrightarrow\underline{\underline{V_3=1\ V}}}$