Question

The power dissipated by $R=2\Omega$ resistor is -

• A. $1\text{W}$
• B. $2\text{W}$
• C. $3\text{W}$
• D. $4\text{W}$

Answer 1

The correct option is B $2\text{W}$

Let us use the technique of successive reduction. The resistor in arm $XQ$ can be removed, as it represents a balanced Wheatstone bridge.



Now arms $PQR$ and $PR$ are in parallel, their equivalent resistance is given by,
$\frac{6\times 3}{6+3}=2\Omega$

$SR$ can be removed as it represents another balanced Wheatstone bridge and equivalent resistance between $PT$ will be,
$\frac{6\times 18}{6+18}=\frac{9}{2}\Omega$
Further, net equivalent resistance,

$R_{\text{eq}}=2+\frac{9}{2}=\frac{13}{2}\Omega$

So, current through $R$, $i=\frac{6.5}{13/2}=1\text{A}$

$\therefore$ Power dissipated,

$P=i^{2}R=1^2\times 2=2\text{W}$