## Application of Kirchhoff’s laws for calculation of total resistance in a circuit

### Task number: 2342

In the picture we can see the schematic of a circuit with 5 resistors. Determine the total resistance between points A and B. Values of resistance for each resistor are shown in the picture.

#### Hint

Look up Kirchhoff’s laws and try to use them.

#### Breakdown

**1.**Let’s say we have connected the circuit in points A, B to the voltage supply*U*(we assume that the voltage supply has internal resistance equal to zero).**2.**At the voltage supply we shall draw an arrow in the direction of electromotive voltage.**3.**In the picture we indicate the currents and choose their direction.**4.**We can choose closed loops and their direction.**5.**Assemble the equations following the first and second Kirchhoff’s laws.**6.**Now we solve the equations.**The more detailed usage of Kirchoff’s laws can be found in assignment Using Kirchhoff’s laws to solve circiut with two power supplies.**#### First part of solution – assembling of the equations

Resistance between points A and B we can calculated with Ohm´s law:

\[U\,=\, R_\mathrm{total}I\,.\]So, we have to calculate total current in the circuit if we connect it to the voltage supply

*U*.From the first Kirchhoff’s law we get equations for 4 knots:

\[I\,=\, I_1 + I_2\,=\, I_4 + I_5\tag{knots A, B}\] \[I_1 + I_3\,=\, I_4\tag{knot C}\] \[I_2\,=\, I_3 + I_5\tag{knot D}\]From the second Kirchhoff’s law we get the following equations:

\[RI_1 - RI_3 - 2RI_2\,=\, 0\,\Rightarrow\,I_1-I_3-2I_2\,=\,0\tag{loop 1}\] \[RI_3 + 2RI_4 - RI_5\,=\, 0\,\Rightarrow\,I_3+2I_4-I_5\,=\,0\tag{loop 2}\] \[U\,=\, 2RI_2 + RI_5\tag{loop 3}\]Now we have 6 equations for 6 unknown currents (

*I*,*I*_{1},*I*_{2},*I*_{3},*I*_{4},*I*_{5}).With this we have completed the solution from the physics point of view. In order to get the final solution we need to solve the simultaneous equations.

#### Second part of the solution – solution for system of linear equations

After we have summed up equations for loop 1 and loop 2 we get equation:

\[I_1 - 2I_2 - I_5 + 2I_4\,=\,0\,.\]In this equation we can substitute currents

*I*_{1}and*I*_{5}from equation (knots A, B):\[I_1\,=\, I - I_2\,,\hspace{20px}I_5\,=\, I - I_4\,,\]

then we get the following equation

\[I - I_2 - 2I_2\,=\, I - I_4 - 2I_4\]therefore we get the following relations

\[I_2\,=\, I_4\hspace{20px}\mathrm{a}\hspace {20px}I_1\,=\, I_5\,.\]

After using these relations, our system of linear equations gets simplified:

\[I\,=\,I_1+I_2\tag{1}\] \[I_1+I_3\,=\,I_2\tag{2}\] \[I_1-I_3-RI_2\,=\,0\tag{3}\] \[U\,=\,2RI_2+RI_1\tag{4}\]Into the third equation (3) we can substitute

\[ I_1 - (I_2-I_1) - 2I_2\,=\, 0 \] \[ 2I_1 - 3I_2\,=\, 0\,.\]*I*_{3}from the second equation (2):and finally we have relation for

\[I_1\,=\, \frac{3}{2}I_2\,.\]*I*_{1}:To calculate total resistance we don´t need the rest of the unknowns, therefore we don´t need to calculate the system of linear equations after this point.

#### Third part of the solution – calculation of total resistance

Now from Ohm´s law we have

\[U\,=\, R_\mathrm{total}I\, =\, R_\mathrm{total}(I_1 + I_2)\]we can substitute

\[R_\mathrm{total}\,=\, \frac{U}{I_1+I_2}\,.\]*R*_{total}from previous equation:From the second Kirchhoff’s law we already know (equation (4) in last paragraph) that

\[U\,=\,2RI_2+RI_1\,,\]and in the previous paragraph we calculated

\[I_1\,=\,\frac{3}{2}I_2\,.\]After we substitute from these equations to the equations for total resistance we get our total resistance

\[R_\mathrm{total}\,=\, \frac{2RI_2+RI_1}{I_1+I_2}\,=\,\frac{2RI_2+\frac{3}{2}RI_2}{\frac{3}{2}I_2+I_2}\] \[R_\mathrm{total}\,=\, \frac{R(2+\frac{3}{2})I_2}{(\frac{3}{2}+1)I_2}\,=\, \frac{7}{5}R\,.\]#### Simplification

Because the circuit is “symmetrical“, there is a known physical reason for currents

*I*_{1}and*I*_{5}to be different. This principle also holds for currents*I*_{2}and*I*_{4}.From Kirchhoff’s laws we immediately get following equations:

\[I\, =\, I_1 + I_2\tag{1.KL, knot A}\] \[I_1+I_3 \,=\, I_2\tag{2.KL, knot C}\] \[RI_1 - RI_3 - 2RI_2 \,=\, 0\tag{loop 2}\] \[U\,=\,2RI_2+RI_1\tag{loop 3}\]And we are already in the second half of the second part of the solution.

#### Answer

Total resistance of the system of resistors given by the first picture between point A and B is \(R_\mathrm{total}\,=\, \frac{7}{5}R\).

#### Similar assignment

Using Kirchhoff’s laws to solve circiut with two power supplies is simpler version of this assignment.

#### Other resolution methods

This assignment can be also solved by transformation of the “triangle“ to a “star“ (to see Transformace trojúhelník - hvězda; in Czech only).