In the figure given below,$A, B$,and $C$ are three ammeters. The ammeter $B$ reads $0.5\, A$. (All the ammeters have negligible resistance.) Calculate:
$(i)$ the readings in the ammeters $A$ and $C$.
$(ii)$ the total resistance of the circuit.

(A) $(i)$ The current in the parallel branches is inversely proportional to the resistance of the branch. The branches containing ammeters $B$ and $C$ have resistances of $6\,\Omega$ and $3\,\Omega$ respectively.
$\frac{\text{Reading of ammeter } C}{\text{Reading of ammeter } B} = \frac{6\,\Omega}{3\,\Omega} = 2$
Therefore,the reading of ammeter $C = 2 \times 0.5\, A = 1.0\, A$.
The reading of ammeter $A$ is the total current in the circuit,which is the sum of currents in the parallel branches:
Reading of ammeter $A = (0.5 + 1.0)\, A = 1.5\, A$.
$(ii)$ Let the total resistance of the circuit be $R$. The circuit consists of a $2\,\Omega$ resistor in series with a parallel combination of $3\,\Omega$ and $6\,\Omega$ resistors.
First,calculate the equivalent resistance of the parallel part $(R_p)$:
$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_p = 2\,\Omega$
Now,the total resistance $R = 2\,\Omega + R_p = 2\,\Omega + 2\,\Omega = 4\,\Omega$.