A capacitor of capacitance C_1 = 1 mu F can w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a series combination,the charge $Q$ on each capacitor is the same.First,calculate the maximum charge each capacitor can hold:$(Q_1)_{max} = C_1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) In a series combination,the charge $Q$ on each capacitor is the same.First,calculate the maximum charge each capacitor can hold:$(Q_1)_{max} = C_1 V_1 = 1\ \mu F 	imes 6\ kV = 6\ \mu C$.$(Q_2)_{max} = C_2 V_2 = 3\ \mu F 	imes 4\ kV = 12\ \mu C$.Since the capacitors are in series,the maximum charge the system can hold is limited by the capacitor with the smaller maximum charge,which is $6\ \mu C$.When the system holds a charge of $Q = 6\ \mu C$,the potential difference across $C_1$ is $V_1 = 6\ kV$ and the potential difference across $C_2$ is $V_2 = Q / C_2 = 6\ \mu C / 3\ \mu F = 2\ kV$.The total maximum voltage the series combination can withstand is $V_{total} = V_1 + V_2 = 6\ kV + 2\ kV = 8\ kV$.

Similar Questions

Three capacitors,each of capacitance $C$ and breakdown voltage $V$,are connected in series. Find the equivalent capacitance and the equivalent breakdown voltage.

The equivalent capacitance between points $A$ and $B$ of the circuit shown will be

Three capacitors each of $4\,\mu F$ are to be connected in such a way that the effective capacitance is $6\,\mu F$. This can be done by connecting them:

Four metallic plates, each with a surface area of one side $A$, are placed at a distance $d$ from each other. The plates are connected as shown in the figure. The capacitance of the system between $a$ and $b$ is

The equivalent capacitance between points $A$ and $B$ in the figure shown below is .......... $\mu F$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module