A current source drives a current in a coil o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the internal resistance of the source be $r$ and its electromotive force be $E$.The current in the coil of resistance $R_1$ is $I_1 = \frac{E}

Vedclass · Accepted Answer

$\sqrt{R_1 R_2}$

Vedclass · Answer

(D) Let the internal resistance of the source be $r$ and its electromotive force be $E$.The current in the coil of resistance $R_1$ is $I_1 = \frac{E}{r + R_1}$.The current in the coil of resistance $R_2$ is $I_2 = \frac{E}{r + R_2}$.Since the heat generated $H = I^2 R t$ is the same for both coils in the same time $t$,we have:$I_1^2 R_1 t = I_2^2 R_2 t$Substituting the expressions for $I_1$ and $I_2$:$\left(\frac{E}{r + R_1}\right)^2 R_1 = \left(\frac{E}{r + R_2}\right)^2 R_2$$\frac{R_1}{(r + R_1)^2} = \frac{R_2}{(r + R_2)^2}$Taking the square root on both sides:$\frac{\sqrt{R_1}}{r + R_1} = \frac{\sqrt{R_2}}{r + R_2}$$\sqrt{R_1}(r + R_2) = \sqrt{R_2}(r + R_1)$$r\sqrt{R_1} + \sqrt{R_1}R_2 = r\sqrt{R_2} + R_1\sqrt{R_2}$$r(\sqrt{R_1} - \sqrt{R_2}) = R_1\sqrt{R_2} - \sqrt{R_1}R_2$$r(\sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_1 R_2}(\sqrt{R_1} - \sqrt{R_2})$$r = \sqrt{R_1 R_2}$

$A$ 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 the heat generated is the same,find the internal resistance of the source.

Similar Questions

If each resistance in the figure is of $9\,\Omega$,then the reading of the ammeter is ............ $A$.

In the circuit shown in the figure,the potential difference between points $A$ and $B$ is........

The potential difference across the $40\,\Omega$ resistor in the given circuit is ............ $V$.

In the network of resistors shown in the adjoining figure,the equivalent resistance between $A$ and $B$ is ............ $ohm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module