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 the $4\,\Omega$ resistor.
$(c)$ Potential difference across the $4\,\Omega$ resistor.
$(d)$ Power dissipated in the $4\,\Omega$ resistor.
$(e)$ Difference in ammeter readings,if any.

(N/A) The two $8\,\Omega$ resistors are in parallel. Their effective resistance $R_p$ is given by:
$R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\,\Omega$.
$(b)$ The total resistance of the circuit $R_{eq} = R_{series} + R_p = 4\,\Omega + 4\,\Omega = 8\,\Omega$. The total current $I$ flowing through 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 current flowing through it is $1\,A$.
$(c)$ The potential difference $V$ across the $4\,\Omega$ resistor is $V = I \times R = 1\,A \times 4\,\Omega = 4\,V$.
$(d)$ The power dissipated $P$ 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 same amount of current flows through every component.