Four resistors are connected to form a quadrilateral ABCD such that...

The given circuit is a Wheatstone bridge with the battery connected across junctions A and C.

The resistances of the arms are $R_{AB} = 10 \, \Omega$, $R_{BC} = 20 \, \Omega$, $R_{AD} = 30 \, \Omega$, and $R_{CD} = 60 \, \Omega$.

Checking the balance condition of the Wheatstone bridge:

$\dfrac{R_{AB}}{R_{BC}} = \dfrac{10}{20} = \dfrac{1}{2}$

$\dfrac{R_{AD}}{R_{CD}} = \dfrac{30}{60} = \dfrac{1}{2}$

Since $\dfrac{R_{AB}}{R_{BC}} = \dfrac{R_{AD}}{R_{CD}}$, the bridge is balanced. Therefore, no current flows through the $50 \, \Omega$ resistor connected between junctions B and D.

The equivalent resistance of the upper branch ABC is $R_{ABC} = 10 + 20 = 30 \, \Omega$.

The equivalent resistance of the lower branch ADC is $R_{ADC} = 30 + 60 = 90 \, \Omega$.

The equivalent resistance of the bridge $R_{eq}$ is the parallel combination of $R_{ABC}$ and $R_{ADC}$:

$R_{eq} = \dfrac{30 \times 90}{30 + 90} = \dfrac{2700}{120} = 22.5 \, \Omega$

The total resistance of the circuit, including the internal resistance of the battery $r = 1.5 \, \Omega$, is:

$R_{total} = R_{eq} + r = 22.5 + 1.5 = 24 \, \Omega$

The current drawn from the battery is:

$I = \dfrac{V}{R_{total}} = \dfrac{12}{24} = 0.5 \, A$

Answer: $0.5 A$