Question

Find the current measured by the ammeter in the circuit shown in the figure (32-E18).

Answer 1

From the symmetry of the circuit,we can see that the potential at point C will be equal to the potential at point F. This implies that no current will flow in this branch. Hence, this arm of the circuit can be removed.
Similarly, the potential at point B will be equal to the potential at point G. So, this arm of the circuit can also be removed.
The simplified circuit diagram will be as shown below.
(b)

From the figure, the effective resistance of the circuit,
$R_{\mathrm{eff}}=\frac{30\times 30}{30+30}=15\mathrm{\Omega }$
The current through the circuit,
$i=\frac{6}{15}=0.4\mathrm{A}$
So, the ammeter reading will be 0.4 A.