AP

Week 6 Electrical & Electronics (Circuit Theory 6)

Introduction

University of SouthamptonCourse: Electricity & Electronics Circuit Theory (6)Instructor: Dr. Josh RobertsonContact: J.J.Robertson@soton.ac.uk

Learning Objectives

  • Understand the basic principle of CR (Capacitor-Resistor) circuits.

  • Understand and apply the transient formulas for charging a capacitor, including the significance of time constants.

Capacitors in Parallel - Recap

  • For the circuit shown:a) Calculate total capacitance using the formula: ( C_{total} = C_1 + C_2 + C_3 + ... )b) Determine charge on each capacitor, considering ( Q = CV ) for each.c) Find total charge held by the circuit as ( Q_T = Q_1 + Q_2 + Q_3 + ... )d) Calculate total energy stored by the circuit using ( E_{total} = \frac{1}{2} C_{total} V^2 ).

Charging Process of Capacitors and Resistors in Series

  • Charge on a capacitor cannot change instantaneously due to its ability to store energy.

  • Current relationship:( I = \frac{dQ}{dt} )

    • As time approaches zero, any change in charge approaches zero.

  • Steady-state current into a capacitor is zero which indicates charge builds up to a point where the voltage across the capacitor equals that of the voltage supply:( V_c = V ).

Step 1 Charging Process

When the switch (S) is closed:

  • Apply Kirchhoff’s Voltage Law (KVL):( E = V_c + V_r )

    • Where ( E ) is the voltage from the supply, ( V_c ) is the voltage across the capacitor, and ( V_r ) is the voltage across the resistor.

Voltage Relationships in Charging

  • The battery voltage (E) remains constant during the charging process.

  • Capacitor voltage can be expressed as:( V_c = \frac{Q}{C} )

  • The voltage drop across the resistor can be calculated:( V_r = IR )

  • KVL yields:( E = \frac{Q}{C} + IR ).

Step 2 - Initial Condition

  • At the moment the switch is closed:

  • Initial charge is ( Q = 0 ) leading to ( V_c = 0 )

  • Initial current through the circuit is given by:( I = \frac{E}{R} ).

Step 3 - After Initial Charge

  • Short time after S is closed (say t1 seconds):

  • Charge partly rises to ( Q_1 ) coulombs.

  • Capacitor voltage is now:( V_{C1} = \frac{Q_1}{C} )

  • Voltage drop across the resistor is now given by ( V_r = I_{t1}R ).

  • Applying KVL at t1:( E = V_{C1} + V_r ).

Step 4 - Continued Charging

  • The charge increases to ( Q_2 ) coulombs at time ( t2 ).

  • Capacitor voltage becomes ( V_C = \frac{Q_2}{C} ).

  • The voltage drop is given by:( V_r = I_{t2}R ).

  • As time progresses, ( V_C ) continuously increases while current & voltage across R decrease due to the charging nature of capacitors.

Step 5 - Steady State

  • After a few seconds, the capacitor becomes fully charged:

  • Current reduces to zero ( ( I = 0 ) ), while the capacitor maintains voltage PMAX, equal to the supply voltage:( V_C = E ).

  • Steady-state established: KVL becomes effective in the form ( E = \frac{Q}{C} + IR ).

Summary of Charging Process Steps

  • Key equations identified:

    • Initial condition: ( V_C = 0 ) + contributions from R.

    • Continuous rise in capacitor voltage ( ( V_C ) ) while current ( ( I ) ) decreases.

    • Steady state occurs when ( V_C = E ) and current ceases.

Voltage-Time Curve for Capacitors

  • Charging follows an exponential growth curve for voltage over time after initiation:

  • Current and voltage across R follow an exponential decay curve, demonstrating the relationship between voltage, charge, and time.

Time Constant in C-R Circuits

  • With constant DC voltage, the time constant is denoted ( ( \tau = RC ) ) measured in seconds.

  • Example calculation: ( R = 1000 \Omega ) and ( C = 10 \mu F ) gives a time constant of ( \tau = 0.0099s ).

Transient Behavior of C-R Circuits

  • Voltage and current response in capacitors exhibit exponential changes:

    • Voltage across capacitor: ( V_C = E(1 - e^{-t/\tau}) ).

    • Voltage drop across resistor: ( V_R = V e^{-t/\tau} ).

    • Current: ( I = I_0 e^{-t/\tau} ) where ( I_0 ) is the initial current.

    • Relationship for resistor current: ( I = \frac{E - V_C}{R} ).

Example Calculation

  • For a 20μF capacitor connected to a 50K resistor with a 20V DC supply:

    • Find the initial current, time constant, current after 1s, voltage across the capacitor after 2s, and time required to achieve 15V across the capacitor.

Further Example Calculation

  • Consider a 100K Ohm resistor with a 5μF capacitor, and a 50V DC supply:

    • Tasks: Calculate capacitor voltage after 1s, current after 2s, and time to reach a voltage of 35V.

Learning Objectives for Discharging Capacitors

  • Understand and apply formulas necessary for discharging a capacitor effectively.

Warm-up Question

  • Find:a) The value of the resistor in a circuit with a 0.5μF capacitor and a time constant of 12ms.b) The capacitor voltage after 7ms when connected to a 10V supply.

Formula Booklet Summarized Equations

  • Capacitor relationships are as follows:

    • Voltage across the capacitor: ( V_C = \frac{Q}{C} )

    • Energy stored in the capacitor: ( E_{cap} = \frac{1}{2} C V^2 )

    • Total charge in the system: ( Q_T = C_pV )

      • Where ( C_p = C_1 + C_2 + C_3 ) for capacitors in parallel.

Charging and Discharging Formulas

  • Charging formula:( V_C(t) = E(1 - e^{-t/\tau}) )( I(t) = I_0 e^{-t/\tau} )

  • Discharging formula:( V_C(t) = V_0 e^{-t/\tau} )( I(t) = -\frac{V_C}{R} )

Charging Process Explained

  • The current flow caused by electrons in the capacitor leads to voltage balance between the capacitor and the resistor, illustrating the relationship:( V_R = I R ).

Voltage and Current Charging Curve

  • Graphical representations of capacitor voltage and charging current against time demonstrate expected behavior during charging.

Fully Charged Capacitor Explanation

  • After 5 time constants, the voltage across the capacitor nearly equals the applied voltage (E).

  • Voltage reaches approximately 63.2% of the voltage difference from its initial value in each time constant period.

Discharging Process

  • Upon switching to position B, the capacitor discharges through a resistor, initially producing maximum current flow in the opposite direction with associated exponential voltage decay.

Discharging Current Graph

  • Illustrative representations of discharging voltage and current over time reveal the expected behavior during the discharge phase.

Discharging Equations

  • During discharge, the capacitor voltage decreases according to:( V_C(t) = V_0 e^{-t/\tau} )

    • This assumes full discharge occurs approximately after 5 time constants.

Discharging Calculations Example

  • For a 100V charged capacitor discharging through a 50K resistor, calculate:

    • Capacitor value, time to reach 20V, current after 0.5s, and resistor voltage after 1s.

Practical Application Example

  • For a 20mF capacitor charged to 120V:

    • Calculate time constants and expected voltages after 10 seconds.

Past Paper Question 3 Summary

  • Analyze circuit charging and discharging under given specifications.

    • Investigate maximum voltage during charge and potential reasons for differing discharging durations.

Past Paper Question Insight

  • Detailed solution outlines provided insights on time constants and specific voltage changes during charging and discharging phases.

Past Paper 2021-22

  • Series connection involving specific criteria to evaluate current flow and time constants.