Question

In Fig. 27-53, , ${\mathit{R}}_{\mathbf{2}}\mathbf{=}{\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{50}\mathbf{.0}\mathbf{\text{\hspace{0.17em}Ω}}$, ${\mathit{R}}_{\mathbf{4}}\mathbf{=}\mathbf{75}\mathbf{.0}\mathbf{\text{\hspace{0.17em}Ω}}$, and the ideal battery has emf. $\mathit{\epsilon }\mathbf{=}\mathbf{6}\mathbf{.00}\mathbf{\text{\hspace{0.17em}V}}\mathbf{}$ (a) What is the equivalent resistance? What is $\mathit{i}\mathbf{}$in (b) resistance 1, (c) resistance 2, (d) resistance 3, and (e) resistance4?

Short answer

a) The equivalent resistance is $R_{eq}=119\text{\hspace{0.17em}Ω}$

b) The current in resistance $1$ is $i_1=5.05\times {10}^{-2}\text{\hspace{0.17em}A}$

c) The current in resistance$2$ is $i_2=1.90\times {10}^{-2}\text{\hspace{0.17em}A}$

d) The current in resistance $3$ is $i_1=1.90\times {10}^{-2}\text{\hspace{0.17em}A}$

e) The current in resistance $4$ is $i_1=1.25\times {10}^{-2}\text{\hspace{0.17em}A}$

Step-by-step solution

Given

i) The resistance $1$ is $R_1=100\text{\hspace{0.17em}Ω}$

ii) The resistance $2$ and resistance $3$ are $R_2=R_3=50.0\text{\hspace{0.17em}Ω}$

iii) The resistance $4$ is $R_4=75.0\text{\hspace{0.17em}Ω}$

iv) The emf of the battery is $\epsilon =6.00\text{\hspace{0.17em}V}$

Understanding the concept 

We can use the concept ofjunction law and loop law. The algebraic sum of potential across any electric component and electric potential is zero. The sum of the currents entering the junction must be equal to the sum of the currents leaving the junction. We can use Ohm’s law.

Formula:

$\mathit{R}\mathbf{=}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{4}}}{{\mathbf{R}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{3}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}{\mathbf{R}}_{\mathbf{4}}\mathbf{+}{\mathbf{R}}_{\mathbf{3}}{\mathbf{R}}_{\mathbf{4}}}$

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{R}}_{\mathbf{1}}\mathbf{+}\mathit{R}$

$\mathit{V}\mathbf{=}\mathit{I}\mathit{R}$

(a) Calculate the equivalent resistance

The equivalent resistance:

According to the circuit diagram, the resistances$R_2$,$R_{3,}$and$R_4$are in parallel, and$R$its equivalent resistance. Then

$\begin{array}{c}\frac{1}{R}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\ R=\frac{R_2{R}_3{R}_4}{R_2{R}_3+R_2{R}_4+R_3{R}_4}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}R=\frac{50\text{\hspace{0.17em}Ω}\times 50.0\text{\hspace{0.17em}Ω}\times 75.0\text{\hspace{0.17em}Ω}}{50.0\text{\hspace{0.17em}Ω}\times 50.0\text{\hspace{0.17em}Ω}+50.0\text{\hspace{0.17em}Ω}\times 75.0\text{\hspace{0.17em}Ω}+50.0\text{\hspace{0.17em}Ω}\times 75.0\text{\hspace{0.17em}Ω}}\\ R=18.75\text{\hspace{0.17em}Ω}\end{array}$

The resistance$R$and$R_1$are in series, then the equivalent resistance of the circuit will be

$R_{eq}=R_1+R$

Substitute all the value in the above equation.

$\begin{array}{c}{R}_{eq}=100\text{\hspace{0.17em}Ω}+18.75\text{\hspace{0.17em}Ω}\\ R_{eq}=118.75\text{\hspace{0.17em}Ω}\\ R_{eq}\simeq 119\text{\hspace{0.17em}Ω}\end{array}$

Hence the equivalent resistance is $R_{eq}=119\text{\hspace{0.17em}Ω}$

(b) Calculate  i in resistance 1

The current in resistance 1 :

According to Ohm’s law,

$i_1=\frac{\epsilon }{R_{eq}}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_1=\frac{6.00\text{\hspace{0.17em}V}}{118.8\text{\hspace{0.17em}Ω}}\\ i_1=5.05\times {10}^{-2}\text{\hspace{0.17em}A}\end{array}$

Hence the current in resistance 1 is $i_1=5.05\times {10}^{-2}\text{\hspace{0.17em}A}$

(c) Calculate i  in resistance 2

The current in resistance 2:

Consider loop $ABCD$as

According to the loop law,

$\begin{array}{c}-i_1{R}_1-i_2{R}_2+\epsilon =0\\ i_2=\frac{\epsilon -i_1{R}_1}{R_2}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_3=\frac{6.00\text{\hspace{0.17em}V}-5.05\times {10}^{-2}\text{\hspace{0.17em}A}\times 100\text{\hspace{0.17em}Ω}}{50.0\text{\hspace{0.17em}Ω}}\\ i_3=1.90\times {10}^{-2}\text{\hspace{0.17em}A}\end{array}$

Hence the current in resistance 3 is $i_2=1.90\times {10}^{-2}\text{\hspace{0.17em}A}$

(d) Calculatei  in resistance 3

The current in resistance:

Consider loop$ABCD$ as

According to the loop law,

$\begin{array}{c}-i_1{R}_1-i_3{R}_3+\epsilon =0\\ i_3=\frac{\epsilon -i_1{R}_1}{R_3}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_3=\frac{6.00\text{\hspace{0.17em}V}-5.05\times {10}^{-2}\text{\hspace{0.17em}A}\times 100\text{\hspace{0.17em}Ω}}{50.0\text{\hspace{0.17em}Ω}}\\ i_3=1.90\times {10}^{-2}\text{\hspace{0.17em}A}\end{array}$

Hence the current in resistance 3 is $i_2=1.90\times {10}^{-2}\text{\hspace{0.17em}A}$

(e) Calculate  i in resistance 4

The current in resistance:

According to the junction law,

$\begin{array}{c}-i_1+i_2+i_3+i_4=0\\ i_4=i_1-i_2-i_3\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}{i}_4=5.05\times {10}^{-2}\text{\hspace{0.17em}A}-1.90\times {10}^{-2}\text{\hspace{0.17em}A}-1.90\times {10}^{-2}\text{\hspace{0.17em}A}\\ i_4=1.25\times {10}^{-2}\text{\hspace{0.17em}A}\end{array}$

Hence the current in resistance $4$ is $i_1=1.25\times {10}^{-2}\text{\hspace{0.17em}A}$