Question 18.4: Applying Kirchhoff’s Rules Goal Use Kirchhoff’s rules to fin...
Applying Kirchhoff’s Rules
Goal Use Kirchhoff’s rules to find currents in a circuit with three currents and one battery.
Problem Find the currents in the circuit shown in Figure 18.14 by using Kirchhoff’s rules.
Strategy There are three unknown currents in this circuit, so we must obtain three independent equations, which then can be solved by substitution. We can find the equations with one application of the junction rule and two applications of the loop rule. We choose junction c. (Junction d gives the same equation.) For the loops, we choose the bottom loop and the top loop, both shown by blue arrows, which indicate the direction we are going to traverse the circuit mathematically (not necessarily the direction of the current). The third loop gives an equation that can be obtained by a linear combination of the other two, so it provides no additional information.
Apply the junction rule to point c . I_{1} is directed into the junction, I_{2} and I_{3} are directed out of the junction.
I_{1}=I_{2}+I_{3}
Select the bottom loop, and traverse it clockwise starting at point a, generating an equation with the loop rule:
\begin{array}{r} \Sigma \Delta V=\Delta V_{\text {bat }}+\Delta V_{4.0 \Omega}+\Delta V_{9.0 \Omega}=0 \\ 6.0 \mathrm{~V}-(4.0 \Omega) I_{1}-(9.0 \Omega) I_{3}=0 \end{array}
Select the top loop, and traverse it clockwise from point c. Notice the gain across the 9.0-\Omega resistor, because it is traversed against the direction of the current!
\begin{aligned} & \Sigma \Delta V=\Delta V_{5.0 \Omega}+\Delta V_{9.0 \Omega}=0 \\ & -(5.0 \Omega) I_{2}+(9.0 \Omega) I_{3}=0 \end{aligned}
Rewrite the three equations, rearranging terms and dropping units for the moment, for convenience:
\begin{aligned} & \text { (1) } I_{1}=I_{2}+I_{3} \\ & \text { (2) } 4.0 I_{1}+9.0 I_{3}=6.0 \\ & \text { (3) }-5.0 I_{2}+9.0 I_{3}=0 \end{aligned}
Solve Equation 3 for I_{2} and substitute into Equation 1:
\begin{aligned} & I_{2}=1.8 I_{3} \\ & I_{1}=I_{2}+I_{3}=1.8 I_{3}+I_{3}=2.8 I_{3} \end{aligned}
Substitute the latter expression into Equation 2 and solve for I_{3} :
4.0\left(2.8 I_{3}\right)+9.0 I_{3}=6.0 \rightarrow I_{3}=0.30 \mathrm{~A}
Substitute I_{3} back into Equation 3 to get I_{2} :
-5.0 I_{2}+9.0(0.30 \mathrm{~A})=0 \rightarrow I_{2}=0.54 \mathrm{~A}
Substitute I_{3} into Equation 2 to get I_{1} :
4.0 I_{1}+9.0(0.30 \mathrm{~A})=6.0 \rightarrow I_{1}=0.83 \mathrm{~A}
Remarks Substituting these values back into the original equations verifies that they are correct, with any small discrepancies due to rounding. The problem can also be solved by first combining resistors.