Circuits

The circuit shown can be redrawn to be as the circuit above. Let the voltage between $A$ and $B$ be $V_{AB}$ . Then the voltages at points $C$ and $D$ , $V_{CB}$ and $V_{DB}$ respectively, by voltage division, are:

$\begin{aligned} V_{CB} & = \frac 4{2+4}V_{AB} = \frac 23 V_{AB} \\ V_{DB} & = \frac 6{3+6}V_{AB} = \frac 23 V_{AB} \\ \implies V_{CB} & = V_{DB} \end{aligned}$

Open Circuit: Since there is no potential difference between $C$ and $D$ , there is no current flowing through the $5\Omega$ resistor, which can thus be replaced by an open circuit. Then the equivalent resistance between $A$ and $B$ is:

$\begin{aligned} R_{AB} & = (2+4) || (3+6) = 6 || 9 = \frac {6\times 9}{6+9} = \boxed{\dfrac {18}5 \Omega} \end{aligned}$

Short Circuit: Since points $C$ and $D$ have the same potential, the $5\Omega$ resistor can be considered as a short circuit. Then the equivalent resistance between $A$ and $B$ is:

$\begin{aligned} R_{AB} & = (2||3) + (4||6) = \frac {2 \times 3}{2+3} + \frac {4 \times 6}{4+6} = \frac 65 + \frac {24}{10} = \boxed{\dfrac {18}5 \Omega} \end{aligned}$