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In a circuit, a Capacitor can be connected in series or in parallel fashion. If a set of capacitors were connected in a circuit, the type of capacitor connection deals with the voltage and current values in that network.

Let us observe what happens, when few Capacitors are connected in Series. Let us consider three capacitors with different values, as shown in the figure below.

When the capacitance of a network whose capacitors are in series is considered, the reciprocal of the capacitances of all capacitors, is added to get the reciprocal of the total capacitance. To get this more clearly,

$$\frac{1}{C_{T}}\:\:=\:\:\frac{1}{C_{1}}\:\:+\:\:\frac{1}{C_{2}}\:\:+\:\:\frac{1}{C_{3}}$$

Following the same formula, if simply two capacitors are connected in series, then

$$C_{T}\:\:=\:\:\frac{C_{1}\:\:\times\:\:C_{2}}{C_{1}\:\:+\:\:C_{2}}$$

Where C_{1} is the capacitance across the 1^{st} capacitor, C_{2} is the capacitance across the 2^{nd} capacitor and C_{3} is the capacitance across the 3^{rd} capacitor in the above network.

The voltage across each capacitor depends upon the value of individual capacitances. Which means

$$V_{C1}\:\:=\:\:\frac{Q_{T}}{C_{1}}\:\:V_{C2}\:\:=\:\:\frac{Q_{T}}{C_{2}}\:\:V_{C3}\:\:=\:\:\frac{Q_{T}}{C_{3}}$$

The total voltage across the series capacitors circuit,

$$V_{T}\:\:=\:\:V_{C1}\:\:+\:\:V_{C2}\:\:+\:\:V_{C3}$$

Where V_{c1} is the voltage across the 1^{st} capacitor, V_{c2} is the voltage across the 2^{nd} capacitor and V_{c3} is the voltage across the 3^{rd} capacitor in the above network.

The total amount of Current that flows through a set of Capacitors connected in series is the same at all the points. Therefore the capacitors will store the same amount of charge regardless of their capacitance value.

Current through the network,

$$I\:\:=\:\:I_{1}\:\:=\:\:I_{2}\:\:=\:\:I_{3}$$

Where I_{1} is the current through the 1^{st} capacitor, I_{2} is the current through the 2^{nd} capacitor and I_{3} is the current through the 3^{rd} capacitor in the above network.

As the current is same, the storage of charge is same because any plate of a capacitor gets its charge from the adjacent capacitor and hence capacitors in series will have the same charge.

$$Q_{T}\:\:=\:\:Q_{1}\:\:=\:\:Q_{2}\:\:=\:\:Q_{3}$$

Let us observe what happens, when few capacitors are connected in Parallel. Let us consider three capacitors with different values, as shown in the figure below.

The total Capacitance of the circuit is the equivalent to the sum of the individual capacitances of the capacitors in the network.

$$C_{T}\:\:=\:\:C_{1}\:\:+\:\:C_{2}\:\:+\:\:C_{3}$$

Where C_{1} is the capacitance across the 1^{st} capacitor, C_{2} is the capacitance across the 2^{nd} capacitor and C_{3} is the capacitance across the 3^{rd} capacitor in the above network.

The voltage measured at the end of the circuit is **same** as the voltage across all the capacitors that are connected in a parallel circuit.

$$V_{T}\:\:=\:\:V_{1}\:\:=\:\:V_{2}\:\:=\:\:V_{3}$$

Where V_{c1} is the voltage across the 1^{st} capacitor, V_{c2} is the voltage across the 2^{nd} capacitor and V_{c3} is the voltage across the 3^{rd} capacitor in the above network.

The total current flowing is equal to the sum of the currents flowing through each capacitor connected in the parallel network.

$$I_{T}\:\:=\:\:I_{1}\:\:+\:\:I_{2}\:\:+\:\:I_{3}$$

Where I_{1} is the current through the 1^{st} capacitor, I_{2} is the current through the 2^{nd} capacitor and I_{3} is the current through the 3^{rd} capacitor in the above network.

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