10.3 Kirchhoff's rules  (Page 4/10)

Now we can apply Kirchhoff’s loop rule, using the map in [link] . Starting at point a and moving to point b , the resistor $R_1$ is crossed in the same direction as the current flow $I_1$ , so the potential drop $I_1{R}_1$ is subtracted. Moving from point b to point e , the resistor $R_2$ is crossed in the same direction as the current flow $I_2$ so the potential drop $I_2{R}_2$ is subtracted. Moving from point e to point f , the voltage source $V_1$ is crossed from the negative terminal to the positive terminal, so $V_1$ is added. There are no components between points f and a. The sum of the voltage differences must equal zero:

Finally, we check loop ebcde . We start at point e and move to point b , crossing $R_2$ in the opposite direction as the current flow $I_2$ . The potential drop $I_2{R}_2$ is added. Next, we cross $R_3$ and $R_4$ in the same direction as the current flow $I_3$ and subtract the potential drops $I_3{R}_3$ and $I_3{R}_4.$ Note that the current is the same through resistors $R_3$ and $R_4$ , because they are connected in series. Finally, the voltage source is crossed from the positive terminal to the negative terminal, and the voltage source $V_2$ is subtracted. The sum of these voltage differences equals zero and yields the loop equation

We now have three equations, which we can solve for the three unknowns.

To solve the three equations for the three unknown currents, start by eliminating current $I_2$ . First add Eq. (1) times $R_2$ to Eq. (2). The result is labeled as Eq. (4):

Next, subtract Eq. (3) from Eq. (2). The result is labeled as Eq. (5):

We can solve Eqs. (4) and (5) for current $I_1$ . Adding seven times Eq. (4) and three times Eq. (5) results in $51\text{Ω}{I}_1=153\text{V},$ or $I_1=3.00\text{A}.$ Using Eq. (4) results in $I_3=\mathrm{-2.00}\text{A}.$ Finally, Eq. (1) yields $I_2=I_1-I_3=5.00\text{A}.$ One way to check that the solutions are consistent is to check the power supplied by the voltage sources and the power dissipated by the resistors:

Note that the solution for the current $I_3$ is negative. This is the correct answer, but suggests that the arrow originally drawn in the junction analysis is the direction opposite of conventional current flow. The power supplied by the second voltage source is 58 W and not −58 W.

Calculating current by using kirchhoff’s rules

Find the currents flowing in the circuit in [link] .

Strategy

This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled $I_1,$ $I_2,$ and $I_3$ in the figure, and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h . In the solution, we apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.

Solution

Applying the junction and loop rules yields the following three equations. We have three unknowns, so three equations are required.

$\begin{array}{c}\text{Junction}c:I_1+I_2=I_3.\hfill \\ \text{Loop}abcdefa:I_1\left(R_1+R_4\right)-I_2\left(R_2+R_5+R_6\right)=V_1-V_3.\hfill \\ \text{Loop}cdefc:I_2\left(R_2+R_5+R_6\right)+I_3{R}_3=V_2+V_3.\hfill \end{array}$

Simplify the equations by placing the unknowns on one side of the equations.

$\begin{array}{c}\text{Junction}c:I_1+I_2-I_3=0.\hfill \\ \text{Loop}abcdefa:I_1\left(3\text{Ω}\right)-I_2\left(8\text{Ω}\right)=0.5\text{V}-2.30\text{V}.\hfill \\ \text{Loop}cdefc:I_2\left(8\text{Ω}\right)+I_3\left(1\text{Ω}\right)=0.6\text{V}+2.30\text{V}.\hfill \end{array}$

Simplify the equations. The first loop equation can be simplified by dividing both sides by 3.00. The second loop equation can be simplified by dividing both sides by 6.00.

$\begin{array}{c}\text{Junction}c:I_1+I_2-I_3=0.\hfill \\ \text{Loop}abcdefa:I_1\left(3\text{Ω}\right)-I_2\left(8\text{Ω}\right)=\text{−}1.8\text{V}.\hfill \\ \text{Loop}cdefc:I_2\left(8\text{Ω}\right)+I_3\left(1\text{Ω}\right)=2.9\text{V}.\hfill \end{array}$

The results are

$I_1=0.20\text{A},I_2=0.30\text{A},I_3=0.50\text{A}.$

Significance

A method to check the calculations is to compute the power dissipated by the resistors and the power supplied by the voltage sources:

$\begin{array}{}\\ P_{R_1}=I_1^2{R}_1=0.04\text{W}.\hfill \\ P_{R_2}=I_2^2{R}_2=0.45\text{W}.\hfill \\ P_{R_3}=I_3^2{R}_3=0.25\text{W}.\hfill \\ P_{R_4}=I_1^2{R}_4=0.08\text{W}.\hfill \\ P_{R_5}=I_2^2{R}_5=0.09\text{W}.\hfill \\ P_{R_6}=I_2^2{R}_6=0.18\text{W}.\hfill \\ P_{\text{dissipated}}=1.09\text{W}.\hfill \\ P_{\text{source}}=I_1{V}_1+I_2{V}_3+I_3{V}_2=0.10\text{W}+0.69\text{W}+0.30\text{W}=1.09\text{W}.\hfill \end{array}$

The power supplied equals the power dissipated by the resistors.

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