Question

Question: The ideal battery in Figure (a) has emf $\mathit{\epsilon }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\text{V}}$. Plot 1 in Figure (b) gives the electric potential difference v that can appear across resistor 1 of the circuit versus the current i in that resistor. The scale of the v axis is set by${\mathit{V}}_{\mathbf{s}}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{\text{V}}$ , and the scale of the i axis is set by${\mathit{i}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\text{mA}}$ . Plots 2 and 3 are similar plots for resistors 2 and 3, respectively. What is the current in resistor 2 in

the circuit of Fig. 27-39a?

Short answer

Answer:

The current through the resistor R₂ is $0.82mA$.

Step-by-step solution

Given data:

$$\begin{gathered} \text{Thevoltage,}\epsilon =6.0\text{V} \\ \text{Thevoltage,}{V}_s=18.0\text{V} \\ \text{Thecurrent,}{I}_s=3.00\text{mA} \end{gathered}$$

Understanding the concept:

Use Kirchhoff’s voltage law. In Kirchhoff’s voltage law addition of the voltage in the loop is zero. Also, you can find the individual resistance values from the slope of the graphs given.

Formula:

$V=IR$

The voltage is define by using following formula.

The equivalent resistor of the parallel resistance is define by,

$R_{12}=\frac{R_1{R}_2}{R_1+R_2}$

Calculate the equivalent resistance:

Here, the slope of the voltage vs the current graph is the resistance so,

The resistors are,

$R_1=6000\Omega ,R_2=4000\Omega and{R}_3=2000\Omega$.

Here, R₁ and R₂ are in parallel. So, the equivalent of these two R₁₂ is as follow

$$\begin{gathered} R_{12}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{4000\Omega \times 6000\Omega }{\left(4000+6000\right)\Omega } \\ =2400\Omega \end{gathered}$$

Now, R₁₂ and R₃ are in series. So, the equivalent is R

$$\begin{gathered} R=R_{12}+R_3 \\ =\left(2000+2400\right)\Omega \\ =4400\Omega \end{gathered}$$