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 inaccurately. PROBLEM Find I1,I2, and I3 in Figure 18.16a. STRATEGY Use Kirchhoff's two rules, the junction rule once and the loop rule twice,

Question

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 inaccurately.

PROBLEM Find [latex]I_{1}, I_{2}[/latex], and [latex]I_{3}[/latex] in Figure [latex]18.16 \mathrm{a}[/latex]

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.

Accepted Answer

Apply Kirchhoff's junction rule to junction [latex]c[/latex]. Because of the chosen current directions, [latex]I_{1}[/latex] and [latex]I_{2}[/latex] are directed into the junction and [latex]I_{3}[/latex] is directed out of the junction.

(1) [latex]I_{3}=I_{1}+I_{2}[/latex]

Apply Kirchhoff's loop rule to the loops [latex]a b c d a[/latex] and [latex]b e f c b[/latex]. (Loop aefda gives no new information.) In loop befcb, a positive sign is obtained when the [latex]6.0-\Omega[/latex] resistor is traversed because the direction of the path is opposite the direction of the current [latex]I_{1}[/latex].

(2) Loop abcda: [latex]10 \mathrm{~V}-(6.0 \Omega) I_{1}-(2.0 \Omega) I_{3}=0[/latex]

(3) Loop befcb: [latex]-(4.0 \Omega) I_{2}-14 \mathrm{~V}+(6.0 \Omega) I_{1}-10 \mathrm{~V}=0[/latex]

Using Equation (1), eliminate [latex]I_{3}[/latex] from Equation (2) (ignore units for the moment):

[latex]10-6.0 I_{1}-2.0\left(I_{1}+I_{2} ight)=0[/latex]

(4) [latex]10=8.0 I_{1}+2.0 I_{2}[/latex]

Divide each term in Equation (3) by 2 and rearrange the equation so that the currents are on the right side:

(5) [latex]-12=-3.0 I_{1}+2.0 I_{2}[/latex]

Subtracting Equation (5) from Equation (4) eliminates [latex]I_{2}[/latex] and gives [latex]I_{1}[/latex] :

[latex]22=11 I_{1} \quad ightarrow \quad I_{1}=2.0 \mathrm{~A}[/latex]

Substituting this value of [latex]I_{1}[/latex] into Equation (5) gives [latex]I_{2}[/latex] :

[latex]\begin{aligned}&2.0 I_{2}=3.0 I_{1}-12=3.0(2.0)-12=-6.0 \mathrm{~A} \&I_{2}=-3.0 \mathrm{~A}\end{aligned}[/latex]

Finally, substitute the values found for [latex]I_{1}[/latex] and [latex]I_{2}[/latex] into Equation (1) to obtain [latex]I_{3}[/latex] :

[latex]I_{3}=I_{1}+I_{2}=2.0 \mathrm{~A}-3.0 \mathrm{~A}=-1.0 \mathrm{~A}[/latex]

REMARKS The fact that [latex]I_{2}[/latex] and [latex]I_{3}[/latex] 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 equa tions are linear, and wrong choices result only in a minus sign in the answer.

Question 18.5: ANOTHER APPLICATION OF KIRCHHOFF'S RULES GOAL Find the curre...

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