# Given a circuit with resistors R1 = 10 W, R2 = 15 W, R3 = 20 and voltage, V_b = 12V, see figure, find the currents through all the resistors?

Jan 2, 2017

${i}_{1} = \frac{12}{13} , {i}_{2} = \frac{12}{65} , {i}_{3} = \frac{48}{65}$

#### Explanation:

By Kirchhoff

${i}_{1} = {i}_{2} + {i}_{3}$

Along a loop

${V}_{b} = {R}_{1} {i}_{1} + {R}_{2} {i}_{2}$

Along the other loop

${R}_{3} {i}_{3} - {V}_{b} - {R}_{2} {i}_{2} = 0$

Joining the equations

$\left\{\begin{matrix}{i}_{1} = {i}_{2} + {i}_{3} \\ {V}_{b} = {R}_{1} {i}_{1} + {R}_{2} {i}_{2} \\ {R}_{3} {i}_{3} - {V}_{b} - {R}_{2} {i}_{2} = 0\end{matrix}\right.$

or also

$\left(\begin{matrix}1 & - 1 & - 1 \\ {R}_{1} & {R}_{2} & 0 \\ 0 & - {R}_{2} & {R}_{3}\end{matrix}\right) \left(\begin{matrix}{i}_{1} \\ {i}_{2} \\ {i}_{3}\end{matrix}\right) = \left(\begin{matrix}0 \\ {V}_{b} \\ {V}_{b}\end{matrix}\right)$

Solving for ${i}_{1} , {i}_{2} , {i}_{3}$ we obtain

$\left(\begin{matrix}{i}_{1} = \frac{\left(2 {R}_{2} + {R}_{3}\right) {V}_{b}}{{R}_{2} {R}_{3} + {R}_{1} \left({R}_{2} + {R}_{3}\right)} \\ {i}_{2} = \frac{\left({R}_{3} - {R}_{1}\right) {V}_{b}}{{R}_{2} {R}_{3} + {R}_{1} \left({R}_{2} + {R}_{3}\right)} \\ {i}_{3} = \frac{\left({R}_{1} + 2 {R}_{2}\right) {V}_{b}}{{R}_{2} {R}_{3} + {R}_{1} \left({R}_{2} + {R}_{3}\right)}\end{matrix}\right)$

NOTE:

Supposing that the figures for ${R}_{1} , {R}_{2}$ are relative to maximum allowed dissipation, the problem would be formulated as:

solve for ${i}_{1} , {i}_{2} , {R}_{1} , {R}_{2}$ the system of equations

{ (i_1^2R_1=10), (i_2^2R_2=15), (i_1 = ((2 R_2 + R_3) V_b)/(R_2 R_3 + R_1 (R_2 + R_3))), (i_2 = ((R_3-R_1) V_b)/(R_2 R_3 + R_1 (R_2 + R_3))) :}

obtaining

${i}_{1} = 2.8 , {i}_{2} = 1.78 , {R}_{1} = 1.27 , {R}_{2} = 4.74$