Question

When a cell is connected to either $2\,\Omega$ or $4.5\,\Omega$ resistors, if the power consumption is same in both the cases, then the internal resistance of the cell is:

• A. $1\,\Omega$
• B. $2\,\Omega$
• C. $3\,\Omega$
• D. $4\,\Omega$

Hint:

If the same cell delivers equal power to two different external resistances $R_1$ and $R_2$, then the internal resistance of the cell is the geometric mean of the two resistances: \[ r=\sqrt{R_1R_2}. \] This is a standard result frequently used in JEE and NEET objective questions and can save considerable calculation time.

Answer 1

The Correct Option is C

Step 1: Write the power delivered to a load.
A cell of emf $E$ and internal resistance $r$ driving a load $R$ delivers $P = \dfrac{E^2 R}{(R + r)^2}$.
Step 2: Apply the equal-power condition.
With $R_1 = 2\,\Omega$ and $R_2 = 4.5\,\Omega$ giving the same power, $\dfrac{E^2 R_1}{(R_1 + r)^2} = \dfrac{E^2 R_2}{(R_2 + r)^2}$. Cancelling $E^2$, $\dfrac{R_1}{(R_1 + r)^2} = \dfrac{R_2}{(R_2 + r)^2}$.
Step 3: Cross multiply.
$R_1 (R_2 + r)^2 = R_2 (R_1 + r)^2$.
Step 4: Use the neat shortcut.
This symmetric condition is satisfied when $r = \sqrt{R_1 R_2}$, the geometric mean of the two resistances.
Step 5: Substitute the values.
$r = \sqrt{2\times 4.5} = \sqrt{9} = 3\,\Omega$.
Step 6: Conclude.
The internal resistance of the cell is $3\,\Omega$, which is option (3).
\[ \boxed{r = 3\,\Omega} \]