Question 18.5: Another Application of Kirchhoff’s Rules GOAL Find the curre...
Another Application of Kirchhoff’s Rules
GOAL Find the currents in a circuit with three currents and two batteries when some current directions are chosen wrongly.
PROBLEM Find I_{\mathrm{1}}, I_{\mathrm{2}}, and I_{\mathrm{3}} in Figure 18.15a.
STRATEGY Use Kirchhoff’s two rules, the junction rule once and the loop rule twice, to develop three equations for the three unknown currents. Solve the equations simultaneously.
Apply Kirchhoff’s junction rule to junction c. Because of the chosen current directions, I_{\mathrm{1}} and I_{\mathrm{2}} are directed into the junction and I_{\mathrm{3}} is directed out of the junction.
(1)\quad I_{3}=I_{1}+I_{2}
Apply Kirchhoff’s loop rule to the loops abcda and befcb.
(Loop aefda gives no new information.) In loop befcb, a positive sign is obtained when the 6.0-Ω resistor is traversed because the direction of the path is opposite the direction of the current I_{\mathrm{1}}.
(2)\ \ \mathrm{Loop}\ a b c d a :\ 10\ V-(6.0~\Omega)I_{1}-(2.0\ \Omega)I_{3}=0
(3)\ \ \mathrm{Loop}\ b efc b: ~-(4.0\;\Omega)I_{2}-\;14\;V+\;(6.0\;\Omega)I_{1}-\;10\;V=0
Using Equation (1), eliminate I_{\mathrm{3}} from Equation (2) (ignore units for the moment):
10-6.0I_{1}-2.0(I_{1}+I_{2})=0
{\bf(4)}\quad10=8.0I_{1}+2.0I_{2}
Divide each term in Equation (3) by 2 and rearrange the equation so that the currents are on the right side:
({5})\;\;\;-1{2}=-3.0I_{1}+2.0I_{2}
Subtracting Equation (5) from Equation (4) eliminates I_{\mathrm{2}} and gives I_{\mathrm{1}}:
22=11I_{1}\ \ \rightarrow\ \ I_{1}=\ 2.0\,\mathrm{A}
Substituting this value of I_{\mathrm{1}} into Equation (5) gives I_{\mathrm{2}}:
2.0I_{2}=3.0I_{1}-12=3.0(2.0)\;-\;12=-6.0\;\mathrm{A}
\begin{array}{r l}{I_{2}=-3.0\,{A}}\end{array}
Finally, substitute the values found for I_{\mathrm{1}} and I_{\mathrm{2}} into Equation (1) to obtain I_{\mathrm{3}}:
I_{3}=I_{1}+I_{2}=2.0\ \mathrm{A}-\,3.0\,\mathrm{A}=-\mathrm{l.0\,A}
REMARKS The fact that I_{\mathrm{2}} and I_{\mathrm{3}} are both negative indicates that the wrong directions were chosen for these currents. Nonetheless, the magnitudes are correct. Choosing the right directions of the currents at the outset is unimportant because the equations are linear, and wrong choices result only in a minus sign in the answer.