When electric bulbs of same power, but different marked voltage are connected in series across the power line, their brightness will be :
Medium
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Resistance of bulb $R=\frac{V^{2}}{P}$, Where $V$ is the marked voltage and $P$ is the marked power $P$. is same for both As both bulbs are connceted in series, so current in both of them will be same. So Brightness $\propto R \propto V^{2}$
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Two long coaxial and conducting cylinders of radius $a$ and $b$ are separated by a material of conductivity $\sigma$ and a constant potential difference $V$ is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is -
It is preferable to measure the $e.m.f.$ of a cell by potentiometer than by a voltmeter because of the following possible reasons.
$(i)$ In case of potentiometer, no current flows through the cell.
$(ii)$ The length of the potentiometer allows greater precision.
$(iii)$ Measurement by the potentiometer is quicker.
$(iv)$ The sensitivity of the galvanometer, when using a potentiometer is not relevant.
Which of these reasons are correct?
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In the given figure, battery $E $ is balanced on $55\, cm$ length of potentiometer wire but when a resistance of $10 \,\Omega$ is connected in parallel with the battery then it balances on $50\, cm$ length of the potentiometer wire then internal resistance $r$ of the battery is ............. $\Omega $
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$n$ identical cells are joined in series with its two cells $A$ and $B$ in the loop with reversed polarities. $EMF$ of each shell is $E$ and internal resistance $r$. Potential difference across cell $A$ or $B$ is (here $n > 4$)
Five cells each of emf $E$ and internal resistance $r$ are connected in series. Due to oversight one cell is connected wrongly. The equivalent internal resistance of the combination is ........... $r$