91影视

Kirchoff鈥檚 law state that at any junction the algebraic sum of all currents should be always equal to zero.

The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero

Consider the following circuit, where ${蔚}^{\text{'}}=24.0鈥�V,鈥�R_1=2.00鈥�惟,鈥�R_2=3.00鈥�惟,鈥�and鈥�R_3=7.00鈥�惟$we need to find the emf $蔚$, such that the current flowing in localid="1668332814111" $R_2=\frac{3R_1}{2}and{R}_3=\frac{7R_1}{2}$is = 1.80 A. First apply the loop rule to the left loop(clockwise), to get

${蔚}^{\text{'}}-蔚+I_1{R}_1-I_2{R}_2=0$ (1)

And to the right loop (also clockwise) to get,

$蔚鈭�I_3{R}_3鈭�I_1{R}_1=0$ (2)

And from the junction rule, we can write,

$I_1+I_2=I_3$ (3)

To simplify the problem, we will use the substitutions $R_2=\frac{3R_1}{2}and{R}_3=\frac{7R_1}{2}$

$\left\{\begin{array}{l}{蔚}^{\mathrm{鈥�}}鈭�蔚+{\mathrm{I}}_1{\mathrm{R}}_1鈭�\frac{3{\mathrm{I}}_2{\mathrm{R}}_1}{2}=0\\ 蔚鈭�\frac{7{\mathrm{I}}_3{\mathrm{R}}_1}{2}鈭�{\mathrm{I}}_1{\mathrm{R}}_1=0\\ {\mathrm{I}}_1+{\mathrm{I}}_2={\mathrm{I}}_3\end{array}.\right.$鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�...(4)

Substitute the third equation in (4) into the first one to eliminate $l_2$

$\begin{array}{r}{蔚}^{\mathrm{鈥�}}鈭�蔚+I_1{R}_1鈭�\frac{3R_1}{2}\left(I_3鈭�I_1\right)=0\\ {蔚}^{\mathrm{鈥�}}鈭�蔚+\frac{5I_1{R}_1}{2}鈭�\frac{3I_3{R}_1}{2}=0\end{array}$(5)

Multiply the second equation in (4) by factor 5/2 so we get,

$\frac{5蔚}{2}鈭�\frac{35l_3{R}_1}{4}鈭�\frac{5l_1{R}_1}{2}=0$ (6)

Adding equations (5) and (6), we get,

$$\begin{gathered} {蔚}^{\mathrm{鈥�}}鈭�蔚+\frac{5I_1{R}_1}{2}鈭�\frac{3I_3{R}_1}{2}+\frac{5蔚}{2}鈭�\frac{35I_3{R}_1}{4}鈭�\frac{5I_1{R}_1}{2}=0 \\ {蔚}^{\mathrm{鈥�}}+\frac{3蔚}{2}鈭�\frac{{41}_3{R}_1}{4}=0 \\ 蔚=\frac{{41}_3{R}_1}{6}鈭�\frac{2{蔚}^{\mathrm{鈥�}}}{3} \end{gathered}$$

Substitute with given data in the above expression, and we get,

$$\begin{gathered} 蔚=\frac{4{\mathbb{1}}_3{\mathrm{R}}_1}{6}鈭�\frac{2{蔚}^{\mathrm{鈥�}}}{3} \\ =\frac{41脳1.80\mathrm{A}脳2.00\mathrm{惟}}{6}鈭�\frac{2脳24.0\mathrm{V}}{3} \\ =8.60\mathrm{V} \end{gathered}$$

Hence, the EMF is 8.60 V.