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Two batteries $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ are connected to three resistors as shown below.
If $V_{1}=2 V$ and $V_{2}=0 \mathrm{~V}$, the current $I=3 \mathrm{~mA}$. If $V_{1}=0 \mathrm{~V}$ and $V_{2}=4 V$, the current $I=4 \mathrm{~mA}$. Now, if $\mathrm{V}_{1}=10 \mathrm{~V}$ and $\mathrm{V}_{2}=10 \mathrm{~V}$, the current I will be-
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If $V_{1}=2 V$ and $V_{2}=0 \mathrm{~V}$, the current $I=3 \mathrm{~mA}$. If $V_{1}=0 \mathrm{~V}$ and $V_{2}=4 V$, the current $I=4 \mathrm{~mA}$. Now, if $\mathrm{V}_{1}=10 \mathrm{~V}$ and $\mathrm{V}_{2}=10 \mathrm{~V}$, the current I will be-
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Verified Answer
The correct answer is:
$25 \mathrm{~mA}$
$\mathrm{I}=\frac{\mathrm{V}_{\mathrm{eq}}}{\mathrm{R}+\mathrm{R}_{\mathrm{eq}}}$
In each case $R_{e q} \& R$ is same only $V_{1} \& V_{2}$ is changing $\therefore V_{\text {eq }}$ is changing
$\mathrm{V}_{\mathrm{eq}}=\frac{2 \times \mathrm{R}_{2}+0 \times \mathrm{R}_{1}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \quad\left[\mathrm{~V}_{1}=2, \mathrm{~V}_{2}=0\right]$
$\mathrm{V}_{\mathrm{eq}}=\frac{2 \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
Case $-2 \quad \mathrm{~V}_{\mathrm{eq}}=\frac{4 \mathrm{R}_{1}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \quad\left[\mathrm{~V}_{1}=0, \mathrm{~V}_{2}=4\right]$
$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{3}{4}=\frac{2 \mathrm{R}_{2}}{4 \mathrm{R}_{1}} \quad \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{3}{2}$
Case - $3 \quad \mathrm{~V}_{\mathrm{eq}}=\frac{10 \mathrm{R}_{1}+10 \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{3}{\mathrm{I}^{\prime}}=\frac{2 \mathrm{R}_{2}}{10\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \Rightarrow \frac{3}{\mathrm{I}^{\prime}}=\frac{2 \times 1.5 \mathrm{R}_{1}}{10\left(2.5 \mathrm{R}_{1}\right)}$ or $\mathrm{I}^{\prime}=25 \mathrm{~mA}$
$\frac{3}{\mathrm{I}^{\prime}}=\frac{2 \mathrm{R}_{2}}{10\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \Rightarrow \frac{3}{\mathrm{I}^{\prime}}=\frac{2 \times 1.5 \mathrm{R}_{1}}{10\left(2.5 \mathrm{R}_{1}\right)}$ or $\mathrm{I}^{\prime}=25 \mathrm{~mA}$
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