Find out the following in the circuit given in the figure.

(a) Two 80 resistors are in parallel, so their effective resistance is given by

$\frac{1}{\mathrm{R}}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ or $\mathrm{R}=4 \Omega$

(b) Net resistance of the circuit, $\mathrm{R}=4 \Omega+4 \Omega=8 \Omega$

$V=8 V$

$\therefore$ Current in the ciruit, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{8}{8}=1 \mathrm{~A}$

Hence current flowing through $4 \Omega$ resistor $=1 \mathrm{~A}$

(c) Potential difference across $4 \Omega$ resistor, $\mathrm{V}=\mathrm{IR}$

$=1 \times 4=4 \mathrm{~V}$

(d) Power dissipated in $4 \Omega$ resistor, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$

$=1 \times 4=4 W$

(e) Since same current flows through every part in a series circuit and both the ammeters are connected in series, so there will be no difference in ammeter readings.