Question

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?

Short answer

Answer

The emf of the ideal battery is 48.3 V .

Step-by-step solution

Given

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

 Step 2: Determine the concept

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:

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

Calculate the current through  

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

Calculate the potential across R2 

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

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

Calculate the current through  R3 

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

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

Calculate the voltage across  R1

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:

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

Calculate the emf of the battery

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