Question

The power developed in an unknown resistance $R$ is the same as the power developed in the resistance of $8\;\Omega$, when these are connected separately across a battery of internal resistance $4\;\Omega$. The value of $R$ is

Answer 1

To solve this question we will use the power developed formula in resistance in terms of current flowing through it and Voltage across it. Then we will equate the power developed in both resistance.

Complete step by step answer:

Consider the first case where it is given that a resistance of $8\;\Omega$ is connected to a battery of internal resistance $4\;\Omega$.The equation for the current flowing through the circuit formed by connecting a resistance of ${R_1}$ to the battery can be written as

${i_1} = \dfrac{V}{{r + {R_1}}}$ Here ${i_1}$ is the current flowing through the battery with internal resistance $r$ when a resistance of ${R_1}$ is connected to it. $V$ is the voltage of the battery. Let’s substitute $8\;\Omega$ for ${R_1}$ and $4\;\Omega$ for $r$ in the equation for current. ${i_1} = \dfrac{V}{{4\;\Omega + 8\;\Omega }}\\\ \Rightarrow {i_1} = \dfrac{V}{{12\;\Omega }}$ Hence, we obtained the equation for current. Now, we can write the relation for the power developed in the resistance $8\;\Omega$ as ${P_1} = {i_1}^2R_1$ Here ${P_1}$ is the power developed in the resistance $8\;\Omega$. Now, we can substitute $\dfrac{V}{{12\;\Omega }}$ for ${i_1}$ in the above equation. Hence, we get $P_1={\dfrac{V}{12}}^2R_1$ Consider the second case where it is given that an unknown resistance $R$ is connected to a battery of internal resistance $4\;\Omega$.The equation for the current flowing through the circuit formed by connecting an unknown resistance $R$ to the battery can be written as

$i = \dfrac{V}{{r + R}}$

Here $i$ is the current flowing through the circuit when a resistance $R$ is connected to the battery.

Let us substitute $4\;\Omega$ for $r$ in the above equation to get,

$i = \dfrac{V}{{\left( {4\;\Omega } \right) + R}}$

Now, we can write the relation for the power developed in the resistance $R$ as

${P_2} = {i}^2R$

Here ${P_2}$ is the power developed in the resistance $R$.

We can substitute $\dfrac{V}{{\left( {4\;\Omega } \right) + R}}$ for $i$ in the above equation for power to get,

$P_2={\dfrac{V}{R+4}}^2R$

It is given in the question that the power developed in the two cases are the same.

Hence, we can write,

$\dfrac{V^2}{144} \times 8$ = $\dfrac{V^2}{(R+4)^2} \times R$

$\dfrac{(R+4)^2}{R}$ = $\dfrac{144}{8}$

$16+R^2+8R$ = $18R$

$R^2-10R+16$

$R^2-8R-2R+16$

$R(R-8)-2(R-8)$

$R=8$ or $R=2$

$\therefore$ The unknown Resistance $R = 2\Omega$ or $R = 8\Omega$.

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Note: There are different types of equations to find the power developed such as equations containing voltage and current only, voltage and resistance only and current and resistance only. We have to choose the power equation wisely so as to get the answer to the question easily.