Question: In Figure, the current in resistance 6 isi₆ =1.40 Aand the resistances are ${\mathbf{R}}_{\mathbf{1}}\mathbf{=}{\mathbf{R}}_{\mathbf{2}}\mathbf{=}{\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\Omega }\mathbf{,}{\mathbf{R}}_{\mathbf{4}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{\Omega },{\mathbf{R}}_{\mathbf{5}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}{\mathbf{\Omega andR}}_{\mathbf{6}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\Omega }\mathbf{.}$What is the emf of the ideal battery?

1. Current $I_6=I_5=1.40A$
2. Resistances $R_1=R_2=R_3=2.00\Omega ,R_4=16.0\Omega ,R_5=8.00\Omega ,R_6=4.00\Omega$

Use the Ohm’s law and Kirchhoff’s loop rule and the junction rule to find the emf of the ideal battery.

Write the formula for the voltage and the current:

$$\begin{gathered} V=IR \\ \sum V=0 \\ \sum I_i=\sum I_o \end{gathered}$$

The total voltage across R₄ is equal to the sum of the voltages across R₅ and R₆ .

Therefore,

$V_4=I_5{R}_5+I_6{R}_6$

$$\begin{gathered} V_4=\left(1.40\text{A}\right)\left(8.00\Omega \right)+\left(1.40\text{A}\right)\left(4.00\Omega \right) \\ {V}_4=16.8V \end{gathered}$$

The current through R₄ is then calculated as:

role="math" localid="1662964664780" $$\begin{gathered} I_4=\frac{V_4}{R_4} \\ {I}_4=\frac{16.8\text{V}}{16.0\Omega } \\ {I}_4=1.05\text{A} \end{gathered}$$

According to the junction rule, the current in R₂ is

$$\begin{gathered} I_2=I_4+I_5 \\ {I}_2=\left(1.05\text{A}\right)+\left(1.4\text{A}\right) \\ {I}_2=2.45\text{A} \end{gathered}$$

So, its voltage is

$$\begin{gathered} V_2=I_2{R}_2 \\ {V}_2=\left(2.45\text{A}\right)\left(2.00\Omega \right) \\ {V}_2=4.90\text{V} \end{gathered}$$

By the loop rule, the voltage across R₃ is

$$\begin{gathered} V_3=V_2+V_4 \\ {V}_3=4.90\text{V}+16.8\text{V} \\ {V}_3=21.7\text{V} \end{gathered}$$

So, the current through R₃ is

$$\begin{gathered} I_3=\frac{V_3}{R_3} \\ {I}_3=\frac{21.7\text{V}}{2.0\Omega } \\ {I}_3=10.85\text{A} \end{gathered}$$

Now, by applying the junction rule, we can get the current through R₁

$I_1=I_2+I_3$

Substitute the values and solve as:

$$\begin{gathered} I_1=2.45\text{A}+10.85\text{A} \\ {I}_1=13.3\text{A} \end{gathered}$$

Therefore, the voltage across R1 is as follows:

$V_1=I_1{R}_1$

Solve further as:

$$\begin{gathered} V_1=\left(13.3\text{A}\right)\left(2.00\Omega \right) \\ {V}_1=26.6\text{V} \end{gathered}$$

Now, by the loop rule,

$\epsilon =V_1+V_3$

Substitute the values and solve as:

$$\begin{gathered} \epsilon =26.6\text{V}+21.7\text{V} \\ \epsilon =48.3\text{V} \end{gathered}$$

Therefore, the emf of the ideal battery is 48.3 V