Question

Question: In Figure,${\mathit{R}}_{\mathbf{1}}\mathbf{=}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathit{\Omega }$and${\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{\Omega }$ . Find the equivalent resistance between points D and E.

(Hint: Imagine that a battery is connected across those points.)

Short answer

Answer:

The equivalent resistance is $R_{eq}=4.50\Omega$.

Step-by-step solution

Given data:

The resistance $R_1=R_2=4.00\Omega$

The resistance $R_3=2.50\Omega$

Understanding the concept:

Use the concept of the equivalent resistance of the series and the parallel circuit. Using these equations, we can find the equivalent resistance of the given circuit.

For series:

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{R}}_{\mathbf{1}}\mathbf{+}{\mathit{R}}_{\mathbf{2}}$

For parallel:

$\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{e}\mathbf{q}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}$

Calculate the equivalent resistance between points D and E.  

The resistor R₁ and R₂ are parallel to each other, so you can find equivalent resistance for that, you get

$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ =\frac{1}{4\Omega }+\frac{1}{4\Omega } \\ =\frac{2}{4}{\Omega }^{-1} \\ =\frac{1}{2}{\Omega }^{-1} \\ R=2\Omega \end{gathered}$$

Now, this equivalent resistance is in series with R₃, so you can write,

$$\begin{gathered} R_{eq}=R+R_3 \\ =2.00\Omega +2.50\Omega \\ =4.50\Omega \end{gathered}$$

Hence, the equivalent resistance is $4.50\Omega$.