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# Three capacitors $3\mu F$ ,$10\mu F$ and $15\mu F$ are connected in series to a voltage source of $100V$ .The charge on $15\mu F$is :A. $22\mu C$B. $100\mu C$C. $2800\mu C$D. $200\mu C$

Last updated date: 21st Jul 2024
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Hint: Capacitance is defined as the ratio of the amount of electric charge deposited on a conductor to the difference in electric potential. Self capacitance and reciprocal capacitance are two capacitance terms that are strongly linked. Any substance that can be electrically charged has a property called self capacitance.

Capacitors are said to be in series as they are connected one after the other. The cumulative capacitance of capacitors in series can be calculated by adding the reciprocals of the individual capacitances and taking the reciprocal of the amount. The overall capacitance is lower than the individual capacitances of the series capacitors since they are connected in series. When two or more capacitors are wired in series, the number of the plate spacings of the individual capacitors results in a single (equivalent) capacitor. Now, coming to the given question;
$\dfrac{1}{{{C_{ef}}}} = \dfrac{1}{3} + \dfrac{1}{{10}} + \dfrac{1}{{15}}$
$\Rightarrow \dfrac{1}{{{C_{ef}}}} = \dfrac{1}{3} + \dfrac{1}{{30}}$
$\Rightarrow \dfrac{1}{{{C_{ef}}}} = \dfrac{3}{6} = \dfrac{1}{2}$
$\therefore {C_{eff}} = 2$
$\therefore$ Charge of ${C_{eff}}$ is $q = {C_{eff}} \times V$
$q = \left( 2 \right) \times \left( {100} \right)$
$q = 200\mu C$
In series all the capacitors have same charge
$\therefore$ Charge on $15\mu F$ is $200\mu C$.
So, the correct option is : (D) $200\mu C$

Note: The capacitance, C, has no negative units and always has a positive value. However, since the Farad is a very broad unit of measurement on its own, sub-multiples of the Farad, such as microfarads, nano-farads, and pico-farads, are commonly used.