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# Three capacitors each of capacitance $C$ are connected to a battery of voltage $V$ as shown in the given figure below. When the key $K$ is closed, the charge which will flow through the battery is: A. $2CV$B. $CV$C. $\dfrac{{CV}}{2}$D. $\dfrac{{CV}}{3}$

Last updated date: 14th Jul 2024
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Hint: In electrical devices, capacitors are the devices which are used to store charges and when Direct current starts to flow across them, they start to getting charged in this process the circuit which completes the capacitors connections gets short circuited and hence no current flows across them.

In the given circuit we can see that, as soon as the key $K$ is closed and when battery starts to allowing the current pass through the whole circuit, the upper two capacitors starts to get charging since, these both capacitors are connected in parallel and this whole mini circuit of upper two capacitors will get short circuited.

Hence, no current or charge will flow across the upper two capacitors and only one lower capacitor of capacitance $C$ will work in the given circuit. As we know, the relation between charge flowing across the capacitor of capacitance $C$ and voltage across the capacitor V is given as:
$Q = CV$
So, the charge flows through the battery for only one lower capacitor.

Hence, the correct option is B.

Note:It must always be checked whether in the given electrical circuits where the capacitors are making short circuited and eliminate that part of circuit while calculating current or charges across the circuit. The parallel combination of two capacitors are calculated as ${C_{net}} = {C_1} + {C_2}$.