ln Fig. E26.11, R1 = 3.00 Ω, R2 = 6.00 Ω, and R3 = 5.00 Ω. The battery has negligible internal resistance. The current I₂ through R₂ is 4.00 A.

(a) What are the currents I₁ and I₃?

(b) What is the emf of the battery?

• R₁ = 3.00 Ω,
• R₂ = 6.00 Ω,
• R₃ = 5.00 Ω
• I₂= 4.00 A

a)

The two resistors R₁ and R₂ are in parallel and the potential difference V across resistors connected in the parallel is the same for every resistor and equals the potential difference across the combination and use the value of the voltage V to find the current across R₁ by using Ohm's law as:

$\mathit{l}\mathbf{=}\frac{\mathbf{\hspace{0.17em}}\mathbf{V}}{\mathbf{R}}$

Voltage drop :

$$\begin{gathered} V_1=V_2 \\ {l}_1{R}_1=l_2{R}_2 \\ {l}_1=\frac{l_2{R}_2}{R_1} \\ =\frac{\left(4.0A\right)\left(6.0\Omega \right)}{3.0\Omega } \\ =8.0A \end{gathered}$$

The combination R₁₂ is in series with R₃, where for the resistors in series, the current is the same therefore I₁₂ = I₃. And as R₁ and R₂ are in parallel, therefore

$I_{12}=I_1+I_2$

And

$$\begin{gathered} l_3=l_1+l_2 \\ =4.0A+8.0A \\ =12.0A \end{gathered}$$

Therefore, the currents I₁ and I₃ are 8.0 A and 12 A respectively.

b)

Find the equivalent resistance in the circuit. R₁ and R₂, are in parallel by the equation as:

$$\begin{gathered} {\mathit{R}}_{\mathbf{12}}\mathbf{=}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}} \\ =\frac{\left(3.0\mathrm{\Omega }\right)\left(6.0\mathrm{\Omega }\right)}{3.0\mathrm{\Omega }+6.0\mathrm{\Omega }} \\ =2.0\mathrm{\Omega } \end{gathered}$$

Also, the combination R₁₂ is in series with R₃ so the equivalent resistance R_eq in the entire circuit is

$$\begin{gathered} R_{eq}=R_{12}+R_3 \\ =2.0\mathrm{\Omega }+5.0\mathrm{\Omega } \\ =7.0\mathrm{\Omega } \end{gathered}$$

The current through resistors connected in series is the same through every resistor and equals the current through the combination, so the emf of the battery in the case of r = 0 is :

$$\begin{gathered} \epsilon =l{R}_{eq} \\ =\left(12.0A\right)\left(7.0\Omega \right) \\ =84.0V \end{gathered}$$

Therefore, the emf of the battery is 84 V.