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Physics-

General

Easy

Question

In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become $left parenthesis e equals 2.718 right parenthesis$

$\frac{10}{e} C$
$\frac{e}{10} C$
$\frac{10}{e^{2}} C$
$\frac{e^{2}}{10} C$

The correct answer is: $fraction numerator 10 over denominator e end fraction C$

Final charge on capacitor is
$q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent$
where $q subscript 0 end subscript$ = charge on capacitor at $t equals 0 comma$
$R C$ =time constant of the circuit.
Putting $q subscript 0 end subscript equals C V subscript 0 end subscript$
$therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript$
Given, $C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s$
Hence, $q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent$
$equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C$

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