R-C Circuits - Charging of Capacitors

An R-C circuit consists of a resistor (R), a capacitor (C), and a voltage source (V) connected in series. When the circuit is closed, the capacitor begins to charge through the resistor. The charging of a capacitor in an R-C circuit is an important concept in physics and electrical engineering, as it describes how the voltage across the capacitor evolves over time.

Charging Process

When a capacitor begins to charge, the current in the circuit is initially high because the voltage across the capacitor is zero. As the capacitor charges, the voltage across it increases, which in turn decreases the current according to Ohm's Law. The charging continues until the voltage across the capacitor equals the voltage of the source, at which point the current becomes zero, and the capacitor is fully charged.

Key Formulas

The voltage across the capacitor as a function of time can be described by the following equation:

[ V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) ]

Where:

• ( V(t) ) is the voltage across the capacitor at time ( t )
• ( V_0 ) is the voltage of the source
• ( R ) is the resistance
• ( C ) is the capacitance
• ( e ) is the base of the natural logarithm (approximately equal to 2.71828)
• ( t ) is the time in seconds

The current in the circuit as a function of time is given by:

[ I(t) = \frac{V_0}{R} e^{-\frac{t}{RC}} ]

The time constant ( \tau ) of an R-C circuit is defined as:

[ \tau = RC ]

The time constant represents the time it takes for the voltage across the capacitor to reach approximately 63.2% of its final value.

Charging Curve

The charging curve of a capacitor is an exponential rise, starting from zero and asymptotically approaching the source voltage ( V_0 ). The time constant ( \tau ) is a measure of how quickly the capacitor charges - a smaller time constant means a faster charge.

Differences and Important Points

Feature	Charging Process	Discharging Process
Initial Voltage	0V (across capacitor)	( V_0 ) (initial voltage across capacitor)
Final Voltage	( V_0 ) (voltage of the source)	0V (across capacitor)
Current Direction	Into the capacitor	Out of the capacitor
Time Constant	( \tau = RC )	( \tau = RC )
Voltage Equation	( V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) )	( V(t) = V_0 e^{-\frac{t}{RC}} )
Current Equation	( I(t) = \frac{V_0}{R} e^{-\frac{t}{RC}} )	( I(t) = -\frac{V_0}{R} e^{-\frac{t}{RC}} )

Examples

Example 1: Charging a Capacitor

A 5V battery is connected to a series circuit containing a 10Ω resistor and a 100μF capacitor. What is the voltage across the capacitor after 1 second?

Given:

• ( V_0 = 5V )
• ( R = 10Ω )
• ( C = 100μF = 100 \times 10^{-6}F )

First, calculate the time constant ( \tau ):

[ \tau = RC = 10Ω \times 100 \times 10^{-6}F = 1 \times 10^{-3}s ]

Now, use the voltage equation to find ( V(t) ):

[ V(t) = 5V \left(1 - e^{-\frac{1s}{1 \times 10^{-3}s}}\right) ] [ V(t) = 5V \left(1 - e^{-1000}\right) ] [ V(t) \approx 5V ]

Since ( e^{-1000} ) is an extremely small number, the voltage across the capacitor after 1 second is approximately equal to the source voltage of 5V.

Example 2: Time to Charge

Using the same circuit as in Example 1, how long does it take for the capacitor to reach 4V?

Given:

• ( V(t) = 4V )
• ( V_0 = 5V )
• ( \tau = 1 \times 10^{-3}s )

Rearrange the voltage equation to solve for ( t ):

[ 4V = 5V \left(1 - e^{-\frac{t}{1 \times 10^{-3}s}}\right) ] [ \frac{4V}{5V} = 1 - e^{-\frac{t}{1 \times 10^{-3}s}} ] [ e^{-\frac{t}{1 \times 10^{-3}s}} = 1 - \frac{4V}{5V} ] [ e^{-\frac{t}{1 \times 10^{-3}s}} = \frac{1V}{5V} ] [ -\frac{t}{1 \times 10^{-3}s} = \ln\left(\frac{1}{5}\right) ] [ t = -1 \times 10^{-3}s \times \ln\left(\frac{1}{5}\right) ]

Calculate the time ( t ):

[ t \approx 3 \times 10^{-3}s ]

It takes approximately 3 milliseconds for the capacitor to reach 4V.

Understanding the charging of capacitors in R-C circuits is crucial for analyzing transient behavior in electrical systems, designing timing circuits, and understanding the dynamics of electric charge in various applications.