Question

A current source drives a current in a coil of resistance ${R_1}$ for a time $t$ . The same source drives current in another coil of resistance ${R_2}$ for the same time. If heat generated is the same, find the internal resistance of the source.
A. $\dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$
B. ${R_1} + {R_2}$
C. $0$
D. $\sqrt {{R_1}{R_2}} $

Answer 1

Hint: The internal resistance of the source will be in series with the external resistance. The current through a resistor having resistance $R$ connected to a voltage source having voltage $V$ is given by $I = \dfrac{V}{R}$ .
The heat generated in the resistor is basically the multiplication of power and time and power is given by ${I^2}R$ where $I$ is the current flowing through the resistance $R$ .

Complete step by step answer:
In this question we have to first find the relation between current flowing through a resistor and voltage given by the source. Ohm’s law given this relation as $V = IR$ . So, the current through a resistor having resistance $R$ connected to a voltage source having voltage $V$ is given by $I = \dfrac{V}{R}$ .
Now, as we know that the internal resistance of the source will be in series with the external resistance. So, the equivalent resistance will be the sum of external resistance and the internal resistance.
Given that the source is the same. Let the emf of the source be $V$ and its internal resistance be $r$ .
Then the current through the first coil is given by
${I_1} = \dfrac{V}{{{R_1} + r}}$
Similarly, the current through the second coil is given by
${I_2} = \dfrac{V}{{{R_2} + r}}$
We know that the heat generated in the resistor is basically the multiplication of power and time and power is given by ${I^2}R$ where $I$ is the current flowing through the resistance $R$ .
Therefore the heat generated in first coil in time $t$ will be
${H_1} = {I_1}^2{R_1}t = {\left( {\dfrac{V}{{{R_1} + r}}} \right)^2}{R_1}t$
Similarly, the heat generated in second coil in same time $t$ will be
${H_2} = {I_2}^2{R_2}t = {\left( {\dfrac{V}{{{R_2} + r}}} \right)^2}{R_2}t$
Now it is given in the question that the heat generated is same for both the coils i.e. ${H_1} = {H_2}$ . So,
${\left( {\dfrac{V}{{{R_1} + r}}} \right)^2}{R_1}t = {\left( {\dfrac{V}{{{R_2} + r}}} \right)^2}{R_2}t$
On simplifying we have
$\dfrac{{{R_1}}}{{{{\left( {{R_1} + r} \right)}^2}}} = \dfrac{{{R_2}}}{{{{\left( {{R_2} + r} \right)}^2}}}$
On cross multiplying we have
${R_1}\left( {{R_2}^2 + {r^2} + 2{R_2}r} \right) = {R_2}\left( {{R_1}^2 + {r^2} + 2{R_1}r} \right)$
On further solving the equation we get
${r^2}\left( {{R_1} - {R_2}} \right) = {R_1}{R_2}\left( {{R_1} - {R_2}} \right)$
But ${R_1} \ne {R_2}$ . So,
$r = \sqrt {{R_1}{R_2}} $ which is the internal resistance of the source.
Hence, option D is correct.

Note: The process of heating of resistance is also known as Joule’s heating or Ohmic heating. In this process, when an electric current passes through a conductor then heat is produced in it.
The main reason for formation of heat in resistors is the collision of the free electrons and many atoms of the resistor when an electric field is applied across a resistor. Mostly this generated heat is dissipated in the air in some time.