# Derivation for voltage across a charging capacitor

The expression obtains the instantaneous voltage across a charging capacitor as a function of time i.e., V (t).

Consider a capacitor connected in series with a resistor, to a constant DC supply through a switch S.

* ‘C’ is the value of capacitance and ‘R’ is the resistance value*. The

**‘V’**is the Voltage of the DC source.

When the switch ‘S’ is closed, the current flows through the capacitor and it acquires the charge. As the capacitor charges, the voltage across the capacitor increases and the current through the circuit gradually decrease. For an uncharged capacitor, the current through the circuit will be maximum at the instant of switching. And the charging currents reaches approximately equal to zero as the potential across the capacitor equals the Source voltage ‘V’.

**Capacitor charging equation derivation steps**

Considering voltage law, the source voltage will be equal to the total voltage drop of the circuit.

Therefore,

Rearrange the equation to perform the integration function,

RHS simplification,

On integrating we get,

As we are considering an uncharged capacitor (zero initial voltage), the value of constant ‘K ‘ can be obtained by substituting the initial conditions of the time and voltage. At the instant of closing the switch, the initial condition of time is t=0 and voltage across the capacitor is v=0.

Thus we get, logV=k for t=0 and v=0.

Take exponential on both sides,

From the above expression, it is clear that the instantaneous voltage will be a result of factors such as capacitance, resistance in series with the capacitor, time and the applied voltage value.

As RC increases, the value of exponential function also increases. That is the rate of voltage rise across the capacitor will be lesser with respect to time. That shows the charging time of the capacitor increase with the increase in the time constant RC.

As the value of time ‘t’ increases, the term reduces and the voltage across the capacitor reaches its saturation value.

## Charge **q** and charging current **i** of a capacitor

The expression for the voltage across a charging capacitor is derived as,

**ν = V(1- e ^{-t/RC})** → equation (1).

**V** – Source voltage

**ν** – instantaneous voltage

**C**– capacitance

**R** – resistance

**t**– time

The Full voltage of a capacitor, **V = Q/C**.

**Q**– Full charge

The instantaneous voltage, **v = q/C**.

**q**– instantaneous charge

**q/C =Q/C** **(1- e ^{-t/RC})**

**q = Q (1- e ^{-t/RC})**

### Charging current

For a capacitor, the flow of the charging current decreases gradually to zero in an exponential decay function with respect to time.

From the voltage law,

**ν = V(1- e ^{-t/RC})**

**ν = V – V e ^{-t/RC}**

**V – ν = V e ^{-t/RC} **→equation(2)

The source voltage, V = voltage drop across the resistor (IR) + voltage across the capacitor ( **ν** ).

**V = i R + ν**

**V – ν = i R**

Substitute **V – ν = i R **in the equation 2.

Therefore, **i R = V e ^{-t/RC}**

** i = (V /R) e ^{-t/RC}**

As V is the source voltage and R is the resistance, **V/R** will be the maximum value of current that can flow through the circuit.

**V/R =Imax**

**i = Imax e ^{-t/RC}**

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