In Figure,${\mathbf{\epsilon }}_1\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{V}$,${\mathbf{\epsilon }}_2\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{V}$ , ${\mathbf{R}}_1\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\Omega }$, ${\mathbf{R}}_1\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\Omega }$ , ${\mathbf{R}}_1\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\Omega }$ and both batteries are ideal. (a) What is the rate at which energy is dissipated in ${\mathbf{R}}_1$ ? (b) What is the rate at which energy is dissipated in ${\mathbf{R}}_2$? (c) What is the rate at which energy is dissipated in ${\mathbf{R}}_3$? (d) What is the power of battery 1? (e) What is the power of battery 2?

i) EMF of 1^st battery, ${ϵ}_1=3.0V$

ii) EMF of 2^nd battery, ${ϵ}_2=1.0V$

iii) Resistor, $R_1=4.0\Omega$

iv) Resistor, $R_2=2.0\Omega$

v) Resistor, $R_3=5.0\Omega$

Use the junction rule separately for each loop. Then conclude the values of currents across each resistor. Using the values of currents, we can calculate the power dissipated in each resistor.

Formula:

Current in block,$i=\frac{V}{R}$

Power dissipated, $P=i^{2}R$

Using the junction rule,

$i_3=i_1+i_2$

Now, let’s consider the loop on the left side.

By the loop rule,

${ϵ}_1-i_1{R}_1-i_3{R}_3=0$

$3.0-4i_1-5(i_1+i_2)=0$ ….. (1)

Similarly, for the loop ontheright side, we can write

${ϵ}_2-i_2{R}_2-i_3{R}_3=0$

${ϵ}_2-i_2{R}_2-(i_1+i_2)R_3=0$

$1-2i_2-5(i_1+i_2)=0$ …… (2)

After solving equations$\left(1\right)$ and $\left(2\right),$ solve as:

$i_1=0.421A$

$i_2=-0.158A$

$i_3=-0.263A$

Power dissipated across resistor$R_1$is calculated as:

$\begin{array}{c}{P}_{R1}=I_1^2{R}_1\\ ={\left(0.421\right)}^2\times 4\\ =0.709\text{\hspace{0.17em}W}\end{array}$

Power dissipated across resistor$R_2$is

$\begin{array}{c}{P}_{R2}=I_2^2{R}_2\\ ={(-0.158)}^2\times 2\\ =0.05\text{\hspace{0.17em}W}\end{array}$

Power dissipated across resistor$R_3$is as follows:

$\begin{array}{c}{P}_{R3}=I_3^2{R}_3\\ ={(-0.263)}^2\times 5\\ =0.346\text{\hspace{0.17em}W}\end{array}$

Power of battery 1 is obtained as follows:

$\begin{array}{c}{P}_{ϵ1}=IV\\ =i_1{ϵ}_1\\ =0.421\times 3\\ =1.26\text{\hspace{0.17em}W}\end{array}$

Power of battery 2 is obtained as follows:

$\begin{array}{c}{P}_{ϵ2}=IV\\ =i_2{ϵ}_2\\ =-0.158\times 1\\ =-0.158\text{\hspace{0.17em}W}\end{array}$