Question

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?

Short answer

(a) The currents I₁ and I₃ are 8.0 A and 12 A respectively.

(b) The emf of the battery is 84 V.

Step-by-step solution

Given data

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

Current in each resisitor

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.

Equivalent resistance

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

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

Emf of the battery

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.