Find out the following in the electric circuit given in the figure:
$(a)$ Effective resistance of two $8\, \Omega$ resistors in the combination.
$(b)$ Current flowing through $4\, \Omega$ resistor.
$(c)$ Potential difference across $4\, \Omega$ resistor.
$(d)$ Power dissipated in $4\, \Omega$ resistor.
$(e)$ Difference in ammeter readings,if any.

(N/A) The two $8\, \Omega$ resistors are connected in parallel. Their effective resistance $R_P$ is given by:
$R_P = \frac{R_1 R_2}{R_1 + R_2} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\, \Omega$
$(b)$ The total resistance of the circuit $R_{eq} = R_{series} + R_P = 4\, \Omega + 4\, \Omega = 8\, \Omega$. The total current $I$ in the circuit is:
$I = \frac{V}{R_{eq}} = \frac{8\, V}{8\, \Omega} = 1\, A$. Since the $4\, \Omega$ resistor is in series with the rest of the circuit,the same current flows through it.
$(c)$ The potential difference $V$ across the $4\, \Omega$ resistor is:
$V = I \times R = 1\, A \times 4\, \Omega = 4\, V$
$(d)$ The power $P$ dissipated in the $4\, \Omega$ resistor is:
$P = I^2 R = (1\, A)^2 \times 4\, \Omega = 4\, W$
$(e)$ There is no difference in the ammeter readings. In a series circuit,the current remains the same at all points,so both ammeters $A_1$ and $A_2$ will show the same reading of $1\, A$.