Question

Someone Plz help in answer this Question.

Answer 1

(a) The combination is parallel therefore, $\frac{1}{8}+\frac{1}{8} = \frac{1}{4} = \frac{1}{R}$. Therefore, resistance will be 4 ohms.

(b) $I= \frac{V}{R}$ = $\frac{8}{4} =2A$

(c) The resistor is connected in series with battery, so the Potential Difference will be 8V

(d) Well $P=VI$ and $V=IR$, therefore $P= \frac{I^2}{R}$
$P= \frac{1^2}{4}=0.25Watt$

(e) Reading for Ammeter 1 = 2A
Reading for Ammeter 2 = 1A