Question

The following circuit diagram Fig. 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 value of R ?

• A. $2\Omega$
• B. $8\Omega$
• C. $6\Omega$
• D. $4\Omega$

Answer 1

The correct option is A $2\Omega$

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 value of R is $5\Omega$
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 ohm.
Hence the combined resistance is given as $\frac{1}{2}+\frac{1}{R}=\frac{1}{1}.$
Therefore the value of R is $2\Omega$
Hence, the value of R is 2 ohms.