Find out the following in the electric circuit given in the figure-

(a) Effective resistance of two 8 ohm resistors in the combination.

(b) Current flowing through the 4-ohm resistor

(a) From the figure, two 8 ohm resistors are connected in parallel combination. Their effective resistance is given by,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{\text R1} + \dfrac {1}{\text R_2}

Here, $\text R$1 = \text R2 = 8 ohm,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{8} + \dfrac {1}{8} \\[1em] = \dfrac {2}{8} \\[1em] \Rightarrow \dfrac {1}{\text R\text P} = \dfrac {1}{4} \\[1em] \Rightarrow \text R_\text P = 4 \text { ohm}

Hence, the effective resistance of two 8 ohm resistors in the combination is 4 ohm.

(b) Since $\text R_\text P$ and 4 ohm resistance are in series, their effective resistance is given by,

$\text R$\text S = \text R\text P + 4 \\[1em] = 4 + 4 \\[1em] \Rightarrow \text R_\text S = 8 \text { ohm}

Let net current flowing through the circuit be '$\text I$'.

Now, this same current flows through $\text R_\text P$ and 4 ohm resistance then

$\text I = \dfrac {\text {Voltage of cell}}{\text R_\text S} \\[1em] \dfrac {8}{8} \\[1em] \Rightarrow \text I = 1\ \text A$

Hence, the current flowing through the 4 ohm resistor is 1 A.