JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 10)
The equivalent resistance of the circuit shown below between points a and b is :
_10th_April_Morning_Shift_en_10_1.png)
16$$\Omega$$
20$$\Omega$$
3.2$$\Omega$$
24$$\Omega$$
Explanation
_10th_April_Morning_Shift_en_10_2.png)
Rearranging the above circuit we get
_10th_April_Morning_Shift_en_10_3.png)
The above circuit is a balanced wheatstone bridge there is no current in the branch $c d$.
_10th_April_Morning_Shift_en_10_4.png)
So we have two $8 \Omega$ and one $16 \Omega$ resistor in parallel. If $R_{\mathrm{eq}}$ is equivalent resistance, then
$$ \begin{aligned} &\frac{1}{R_{\mathrm{eq}}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\frac{5}{16} \\\\ &\Rightarrow R_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega \end{aligned} $$
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