Question 254: Current Electricity

Question

The equivalent resistance between A and B as shown in figure is:

Accepted Answer

Potential across
$\mathrm{R}_1=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}$
$=V_A-V_B$    $\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}ight)$
Potential across
$R_2=V_D-V_C$
$=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$    $\left(\because \mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{A}}ight)$
Potential across
$R_3=V_D-V_B$ $=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$
    $\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}ight)$
It means all the resistance are in parallel
$$
\begin{aligned}
& \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3} \\
& \Rightarrow \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10} \\
& \Rightarrow \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1+1+2}{20} \\
& \Rightarrow \mathrm{R_{eq}}=\frac{20}{4}=5 \mathrm{k} \Omega
\end{aligned}
$$

Answer

$$30 ~\mathrm{k} \Omega$$

Answer

$$20 ~\mathrm{k} \Omega$$

Answer

$$5 ~\mathrm{k} \Omega$$

Answer

$$10 ~\mathrm{k} \Omega$$

Current Electricity

The equivalent resistance between A and B as shown in figure is:

Choose the correct answer: