Question

A cell of constant emf first connected to a resistance \[{R_1}\] and then connected to a resistance\[{R_2}\]. If power delivered in both cases is the same, then the internal resistance of the cell is
A. \[\dfrac{{{R_1} - {R_2}}}{2}\]
B. \[{R_1} + {R_2}\]
C. \[\sqrt {{R_1}{R_2}} \]
D. \[\sqrt {{R_1} + {R_2}/2} \]

Answer 1

Hint: The total resistance of the circuit is the sum of internal resistance of cell and external resistance connected in the circuit. Use Ohm’s law to express the current in the circuit. Since the power in both cases is the same, use expression for power and equate.

Formula used:
\[I = \dfrac{E}{{R + r}}\]
Here, E is the emf of the cell, R is the resistance and r is the internal resistance of the cell.

Complete step by step answer:
The total resistance of the circuit when the resistance \[{R_1}\] is connected will be,
\[R = {R_1} + r\] …… (1)
Here, r is the internal resistance of the cell.

According to Ohm’s law, the current in the circuit will be,
\[I = \dfrac{E}{{{R_1} + r}}\] …… (2)
Here, E is the emf of the cell.

The power delivered to the circuit when the resistance \[{R_1}\] is connected in the circuit is,
\[P = {I^2}R\]
 Use equation (1) and (2) to rewrite the above equation as follows,
\[P = {\left( {\dfrac{E}{{{R_1} + r}}} \right)^2}{R_1}\] …… (3)

The total resistance of the circuit when the resistance \[{R_2}\] is connected will be,
\[R = {R_2} + r\] …… (4)
Here, r is the internal resistance of the cell.

Now, the current in the circuit will be,
\[I = \dfrac{E}{{{R_2} + r}}\] …… (5)

The power delivered to the circuit when the resistance \[{R_2}\] is connected in the circuit is,
\[P = {\left( {\dfrac{E}{{{R_2} + r}}} \right)^2}{R_2}\] …… (6)

Since the power delivered in the both cases is the same, we can write,
\[{\left( {\dfrac{E}{{{R_1} + r}}} \right)^2}{R_1} = {\left( {\dfrac{E}{{{R_2} + r}}} \right)^2}{R_2}\]
\[ \Rightarrow {R_1}{\left( {{R_2} + r} \right)^2} = {\left( {{R_1} + r} \right)^2}{R_2}\]
\[ \Rightarrow {R_1}\left( {R_2^2 + 2{R_2}r + {r^2}} \right) = \left( {R_1^2 + 2{R_1}r + {r^2}} \right){R_2}\]
\[ \Rightarrow {R_1}R_2^2 + 2{R_1}{R_2}r + {R_1}{r^2} = {R_2}R_1^2 + 2{R_2}{R_1}r + {R_2}{r^2}\]
\[ \Rightarrow {R_1}R_2^2 + {R_1}{r^2} = {R_2}R_1^2 + {R_2}{r^2}\]
\[ \Rightarrow {R_1}R_2^2 - {R_2}R_1^2 = {R_2}{r^2} - {R_1}{r^2}\]
\[ \Rightarrow {R_1}{R_2}\left( {{R_2} - {R_1}} \right) = {r^2}\left( {{R_2} - {R_1}} \right)\]
\[ \therefore r = \sqrt {{R_1}{R_2}} \]

So, the correct answer is option (C).

Note: The internal resistance of the cell is the resistance due to electrolytic material of the cell. It is usually negligible, therefore, in calculations it is neglected from the equivalent resistance of the circuit. The power in the electrical circuit is the product of current and voltage. After substituting the expression for current from using Ohm’s law, you will get the desired formula for power in terms of resistance.