When a battery sends current through the resistances R1 and R2R- the internal resistance of battery is -

The correct option is D √(R_1R_2) Here current through the resistance R1 is I_1 = εr+ R_1 and current through the resistance R2 is I_2 = εr+ R_2 . ...

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Byju's Answer

Standard XII

Physics

Non-Ideal Battery

When a batter...

Question

When a battery sends current through the resistances $R_1$ and $R_2$, the internal resistance of battery is :

\sqrt{\frac{R_{1} R_{2}}{2}}

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\sqrt{R_{1} R_{2}}

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\frac{R_{1} - R_{2}}{2}

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\frac{R_{1} + R_{2}}{2}

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Solution

The correct option is D $\sqrt{R_{1} R_{2}}$
Here current through the resistance R1 is
$I_{1} = \frac{ε}{r + R_{1}}$
and current through the resistance R2 is
$I_{2} = \frac{ε}{r + R_{2}}$ .
Also the heat produced due to passage of current through both the resistances is same.
Hence,
$Q = I_{1}^{2} R_{1} = I_{2}^{2} R_{2}$
Using the values of $I_{1}$ and $I_{2}$ in above equation we get
${(\frac{ε}{r + R_{1}})}^{2} R_{1} = {(\frac{ε}{r + R_{2}})}^{2} R_{2}$
where, $ε$ is the emf and $r$ is the internal resistance of battery.
$\frac{ε^{2} R_{1}}{{(r + R_{1})}_{1}^{2}} = \frac{ε^{2} R_{2}}{{(r + R_{2})}_{2}^{2}}$
$r^{2} (R_{1} - R_{2}) = R_{1} R_{2} (R_{1} - R_{2})$
$r^{2} = R_{1} R_{2}$
$∴ r = \sqrt{R_{1} R_{2}}$

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When a battery sends current through the resistances $R_1$ and $R_2R$, the internal resistance of battery is :

When a battery sends current through the resistances $R_1$ and $R_2$, the internal resistance of battery is :