In Fig. 27-70, the ideal battery has emf $\mathbf{\epsilon }\mathbf{=}\mathbf{30}\mathbf{.0}\mathbf{}\mathbf{\text{V}}$, and the resistances are${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{}{\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{14}\mathbf{\text{Ω}}$, ${\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{}{\mathbf{R}}_{\mathbf{4}}\mathbf{=}\mathbf{}{\mathbf{R}}_{\mathbf{5}}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.0}\mathbf{\text{Ω}}$, ${\mathbf{R}}_{\mathbf{6}}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{\text{Ω}}$, and ${\mathbf{R}}_{\mathbf{7}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.5}\mathbf{\text{Ω}}$. What are currents (a)${\mathbf{i}}_{\mathbf{2}}$, (b) ${\mathbf{i}}_{\mathbf{4}}$, (c) ${\mathbf{i}}_{\mathbf{1}}$, (d) ${\mathbf{i}}_{\mathbf{3}}$, and (e)${\mathbf{i}}_{\mathbf{5}}$ ?

Fig. 27-70

1. Emf of the ideal battery,$\mathrm{\epsilon }=30.0\text{V}$
2. The given resistances are:$R_1=R_2=14\text{Ω}$${\mathrm{R}}_3={\mathrm{R}}_4={\mathrm{R}}_5=6\text{Ω}$$R_6=2.0\text{Ω}$${\mathrm{R}}_7=1.5\text{Ω}$

In a circuit with several resistors and an ideal battery, the current flow differs due to many conditions. For resistors parallel to each other, the current flowing through them differs, and that of the current value divides into equal values at the junction. Thus, using the concept of voltage law, the current through each resistor as given in the figure can be written and further calculated following the given data.

Formula:

The voltage equation using Ohm’s law, $\mathbf{V}\mathbf{=}\mathbf{IR}$ (i)

The equivalent resistance for a series combination, ${\mathbf{R}}_{\mathbf{eq}}\mathbf{=}\sum _i^n{R}_n$ (ii)

The equivalent resistance for a parallel combination,${\mathbf{R}}_{\mathbf{eq}}\mathbf{=}\sum _i^n\frac{1}{R_n}$ (iii)

Kirchhoff’s voltage law, $\mathbf{\sum }_{\mathbf{closed}\mathbf{}\mathbf{loop}}\mathbf{V}\mathbf{=}\mathbf{0}$ (iv)

Kirchhoff’s junction rule, ${\mathbf{I}}_{\mathbf{in}}\mathbf{=}{\mathbf{I}}_{\mathbf{out}}$ (v)

(a)

The equivalent resistance of the four resistors on the left side is given using equation (ii) and (iii) as follows:

$\begin{array}{c}{\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_{12}+{\mathrm{R}}_{34}\\ =\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}+\frac{{\mathrm{R}}_3{\mathrm{R}}_4}{{\mathrm{R}}_3+{\mathrm{R}}_4}\\ =\frac{\left(14\text{Ω}\right)\left(14\text{Ω}\right)}{14\text{Ω}+14\text{Ω}}+\frac{\left(\text{6Ω}\right)\left(\text{6Ω}\right)}{\text{6Ω}+6\text{Ω}}\\ =7.0\text{Ω+3.0Ω}\end{array}$

Now, the current value${\mathrm{i}}_2$ can be calculated using the emf of the battery in equation (i) as follows:

$\begin{array}{c}{\mathrm{i}}_2=\frac{30\text{V}}{10\text{Ω}}\\ =3.0\text{A}\end{array}$

Hence, the value of the current is $3.0\text{A}$.

(b)

The combination of the three resistors on the right will give the equivalent resistance using equations (ii) and (iii) as follows:

$\begin{array}{c}{\mathrm{R}}_{\mathrm{eq}}\text{'}={\mathrm{R}}_{56}+{\mathrm{R}}_7\\ =\frac{{\mathrm{R}}_5{\mathrm{R}}_6}{{\mathrm{R}}_5+{\mathrm{R}}_6}+{\mathrm{R}}_7\\ =\frac{\left(\text{6Ω}\right)\left(\text{2Ω}\right)}{\text{6Ω}+2\text{Ω}}+1.5\text{Ω}\\ =3.0\text{Ω}\end{array}$

Now, the current value$i_4$ can be calculated using the emf of the battery in equation (i) as follows:

$\begin{array}{c}{\mathrm{i}}_4=\frac{30\text{V}}{\text{3Ω}}\\ =10.0\text{A}\end{array}$

Hence, the value of the current is $10.0\text{A}$.

(c)

By the junction rule that is equation (v), the current value${\mathrm{i}}_1$ can be calculated as follows:

$\begin{array}{c}{\mathrm{i}}_1={\mathrm{i}}_2+{\mathrm{i}}_4\\ =3.0\text{A}+10.0\text{A}\\ =13.0\text{A}\end{array}$

Hence, the value of the current is $13.0\text{A}$.

(d)

By concept of the symmetry, the current value $i_1$can be calculated as follows:

$\begin{array}{c}{\mathrm{i}}_3=\frac{{\mathrm{i}}_2}{2}\\ =\frac{\text{3.0A}}{2}\\ =1.5\text{A}\end{array}$

Hence, the value of the current is $1.5\text{A}$.

(e)

By the loop rule from equation (iv), the current $i_5$ can be calculated as follows:

$\begin{array}{c}\mathrm{\epsilon }-{\mathrm{i}}_4\left({\mathrm{R}}_7\right)-{\mathrm{i}}_5\left({\mathrm{R}}_6\right)=0\\ 30\text{V}-10\text{A}\left(1.5\text{Ω}\right)-{\mathrm{i}}_5\left(2.0\text{Ω}\right)=0\\ {\mathrm{i}}_5=\frac{15\text{V}}{2.0\text{Ω}}\\ =7.5\text{A}\end{array}$

Hence, the value of the current is $7.5\text{A}$.