In the circuit diagram of figure, $E = 5\, volt, r = 1\, \Omega ,$$ R_2 = 4\, \Omega , R_1 = R_3 = 1 \Omega$ and $C = 3\, μF.$ Then the magnitude of the charge on each capacitor plate is......$\mu C$
A$6 $
B$12$
C$24 $
D$0 $
Medium
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A$6 $
a $6 \,\mu \mathrm{C},$ Hint : the $p.d$ across $\mathrm{R}_{2}$ is $4$ $volt.$ The total charge in the two parallel arms is $12\, \mu C$.
Thus in each row,
$q=6\, \mu C.$
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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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