Solution:
(a) Since, two 
resistors are in parallel, then their effective resistance 
is given by
![\[\begin{aligned}\frac{1}{R_{p}} &=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} \S_{p} &=4 \Omega\end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1fc71d9329cbe0e8bd524063f64dc21a_l3.png "Rendered by QuickLaTeX.com")
b)
Total resistance in the circuit.
![\[R=4 \Omega+R_{p}=4 \Omega+4 \Omega=8 \Omega\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a74ebcbfb0c5f639c37c23fcccb0eac7_l3.png "Rendered by QuickLaTeX.com")
Current through the circuit,
![\[I=\frac{V}{R}=\frac{8}{8}=1 A\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0039a9473a96655c9919eb6feb27894b_l3.png "Rendered by QuickLaTeX.com")
Thus, current through
resistor is 
as and are in series and same current flows through them.