Observe the electric circuit below and answer the questions that follow. (a) If we increase the current in the above circuit using a rheostat, will the reading of the voltmeter decrease or increase? (b) Give a reason for answer ‘a’. (c) The present reading of the voltmeter is 2 V, and the current in the circuit is 0.8 A. Calculate the potential drop when the current in the circuit becomes 1 A. Given E = 3 V.

 

(a) Decrease.

(b) This is because the voltmeter shows terminal voltage. Increase in current increases potential drop across internal resistance which decreases terminal voltage.

and Terminal Voltage = Emf - Internal Voltage drop ( Ir)

(c) Internal Voltage drop (Potential Drop) = EMF - Terminal voltage

Initially,

Potential drop = 3 - 2 = 1 V

Let internal resistance = r

$$\begin{gathered} \text{Potential drop}=\text{I}\times \text{r} \\ \Rightarrow \text{r}={\displaystyle \frac{\text{Potential drop}}{\text{I}}} \\ \Rightarrow \text{r}={\displaystyle \frac{1}{0.8}}=1.25\mathrm{\Omega } \end{gathered}$$

$Now,\; Current\; Increased\; to\; 1\; A$

New potential drop = 1 × 1.25 = 1.25 V