91Ó°ÊÓ

Kirchhoff’s junction rule says that the total current into a junction equals the total current out of the junction. This is a statement of conservation of charge. It is also sometimes called Kirchhoff’s first law, Kirchhoff’s current law, the junction rule, or the node rule.

${\mathbf{l}}_{\mathbf{1}}\mathbf{=}{\mathbf{l}}_{\mathbf{2}}\mathbf{+}{\mathbf{l}}_{\mathbf{3}}$ (1)

Rearrange the equation to eliminate the variable in the next step,

${\mathbf{l}}_{\mathbf{2}}\mathbf{=}{\mathbf{l}}_{\mathbf{1}}\mathbf{-}{\mathbf{l}}_{\mathbf{3}}$ (2)

Loop 1,

$$\begin{gathered} 42.0V-R_1{I}_1-R_2{I}_2=0 \\ 42.0V-\left(8.00\mathrm{Î\copyright }\right)I_1-\left(6.00\mathrm{Î\copyright }\right)I_2=0 \end{gathered}$$ (3)

Loop 2,

$$\begin{gathered} 42.0V-R_1{I}_1-R_3{I}_3=0 \\ 42.0V-\left(8.00\mathrm{Î\copyright }\right)I_1-\left(3.00\mathrm{Î\copyright }\right)I_3=0 \end{gathered}$$ (4)

Now, use equations (3) and (4) to estimate the values of current, such that,

$$\begin{gathered} 21.0V-\left(4.00\mathrm{Î\copyright }\right)I_1-\left(3.00\mathrm{Î\copyright }\right)I_2=0 \\ 21.0V-\left(4.00\mathrm{Î\copyright }\right)I_1-\left(3.00\mathrm{Î\copyright }\right)\left(I_1-I_3\right)=0 \\ 21.0V-\left(7.00\mathrm{Î\copyright }\right)I_1+\left(3.00\mathrm{Î\copyright }\right)I_3=0 \end{gathered}$$

Now subtract the above expression, from equation (4), and we get,

$$\begin{gathered} 21.0V-\left(7.00\mathrm{Î\copyright }\right)I_1+\left(3.00\mathrm{Î\copyright }\right)I_3=0 \\ 42.0V-\left(8.00\mathrm{Î\copyright }\right)I_1+\left(3.00\mathrm{Î\copyright }\right)I_3=0 \end{gathered}$$

And therefore,

$$\begin{gathered} 63.0V-\left(15.00\mathrm{Î\copyright }\right)I_1=0 \\ {I}_1=\frac{63.0v}{15.00\mathrm{Î\copyright }} \\ {I}_1=4.2A.......................\left(5\right) \end{gathered}$$

Now substitute the value of I₁from equation (5) to equations (3) and (4),

role="math" localid="1664256430747" $$\begin{gathered} 21.0V-\left(8.00\mathrm{Î\copyright }\right)×4.2A-\left(6.00\mathrm{Î\copyright }\right)I_2=0 \\ {I}_2=\frac{8.4V}{3.00\mathrm{Î\copyright }} \\ {I}_2=1.8A \end{gathered}$$

Further,

$$\begin{gathered} 42.0V-\left(8.00\mathrm{Î\copyright }\right)×4.2A-\left(6.00\mathrm{Î\copyright }\right)I_2=0 \\ {I}_2=\frac{8.4V}{6.00\mathrm{Î\copyright }} \\ {I}_2=1.4A \end{gathered}$$

Now, use the above values to obtain the total current, such that

$$\begin{gathered} V_b-I{R}_1-I{R}_2=0 \\ 42.0V-\left(8.00\mathrm{Î\copyright }\right)×I-\left(6.00\mathrm{Î\copyright }\right)×I=0 \\ I=\frac{42V}{14.00\mathrm{Î\copyright }} \\ I=3.0A \end{gathered}$$

And, now we can write,

$$\begin{gathered} V_b-I{R}_1-0-V_c=0 \\ 42.0V-\left(8.00\mathrm{Î\copyright }\right)×\left(3.00A\right)-V_c=0 \\ {V}_c=18.0V \end{gathered}$$

As we know, that charge on the capacitor can be expressed as,

$$\begin{gathered} Q=C{V}_c \\ =\left(4.0×{10}^{-6}F\right)×18.0V \\ =7.2×{10}^{-5}C \end{gathered}$$

Hence the charge is $7.2×{10}^{-5}C$