MCQ
The figure given below shows three resistors ? Their combined resistance is :
  • A
    $1\frac{5}{7}\Omega$
  • B
    $14\Omega$
  • C
    $6\frac{2}{3}\Omega$
  • $7\frac{1}{2}\Omega$

Answer

Correct option: D.
$7\frac{1}{2}\Omega$
The resistors of $6I$ and $2I$ are connected in parallel.
$\therefore\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$ Here,
$\text{R}_1=6\Omega$
$\text{R}_2=2\Omega\ \frac{1}{\ \text{R}}$
$=\frac{1}{6}+\frac{1}{2}\frac{1}{\text{R}}=\frac{4}{6}\ \text{R}=\frac{6}{4}$
This arrangement is further connected in series with the $6I$ resistor.
$\therefore$ Net resistance $=\frac{6}{4}+6=7\frac{1}{2}\Omega$

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