Question

The circuit diagram shows that resistors $2\Omega$, $4\Omega$ and $R\Omega$ connected to a battery of e.m.f. 2 V and internal resistance $3\Omega$. A main current of 0.25 A flows through the circuit. What is the p.d. across the $R\Omega$ and $2\Omega$ resistor?

• A. 0.25 V, 0.5 V
• B. 0.25 V, 0.25 V
• C. 2 V, 1 V
• D. 0.5 V, 0.25 V

Answer 1

Answer: (B)

0.25 V, 0.25 V

The current in the circuit is calculated from the formula $I=\frac{\epsilon }{R+r}$ where,
$\epsilon$- emf of the cell
R- external resistance
r- internal resistance of the cell.
When the cell of internal resistance of 3 ohms and the total current in the circuit is given as 0.25 A the current relation is given as
That is, $0.25=\frac{2}{R+3}.$
The Resistance is $R=5\Omega$
Hence, the total resistance in the circuit is 5 ohms, out which one resistor is 4 ohms. So, the unknown combined resistance is $5-4=1\Omega .$
The potential difference across the 2 ohms or R resistor is given as $V=IR=0.25A\times 1\Omega =0.25V$
Thus, the p.d. across the R or 2 ohms resistor is 0.25 V.