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W11D1: RC Circuits

Reading Course Notes: Sections 7.7-7.8, 7.11.3, 7.13.5

Announcements

Math Review Week 11 Tuesday 9 pm-11 pm in 26-152 PS 8 due Week 11 Tuesday at 9 pm in boxes outside 32-082 or 26-152 Exam 3 Thursday April 24 7:30 pm –9:30 pm Conflict Exam 3 Friday April 25 8-10 am, 10-12 noon If you have a regularly scheduled MIT activity that conflicts with Exam 3, contact padour@mit.edu and explain conflict and request time to take conflict exam. Sophomore Conflict due to Ring Delivery Event: 8-10 am or 9-11 am in 32-123.

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Outline DC Circuit Rules for Capacitors

RC Charging Circuit First Order Linear Differential Equations RC Discharging Circuit

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Review DC Circuits Rules for Batteries and Resistors

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Reference System for Circuit Analysis

1.  Simplify resistors in series/parallel. 2.  Assign current and current direction in each

branch 3.  Assign circulation direction for each loop. 4.  Circulation direction across each element

defines “before” and “after” side of element. 5.  Define electric potential difference across a

circuit element by

ΔV ≡V (after)−V (before) ≡V (a)−V (b)

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Sign Conventions - Battery Circulating from the negative to positive terminal of a battery increases your potential

ΔV = +ε

Circulating from the positive to negative terminal of a battery decreases your potential

ΔV = −ε

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Sign Conventions - Resistor Direction of current is the same as direction of electric field which points to lower electric potential

Circulating across a resistor in the direction of current decreases the potential

Circulating across a resistor in the opposite direction of current increases your potential

ΔV = − IR

ΔV = + IR

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DC Circuits Rules for Charged Capacitors and Current

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Sign Conventions – Charged Capacitor and Current

1.  Assign and to each side of a capacitor

I = + dQ

dt

I = − dQ

dt

+Q −Q2. When direction of current is toward the positive plate of a capacitor, then 3. When direction of current is away from the positive plate of a capacitor, then

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Sign Conventions - Capacitor Circulating across a capacitor from the negatively to positively charged plate increases the electric potential

ΔV = +Q / C

Circulating across a capacitor from positively charged plate to negatively charged plate decreases the electric potential

ΔV = −Q / C

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Kirchhoff’s Loop Rules

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Charge Conservation Sum of currents entering any junction in a circuit must equal sum of currents leaving that junction.

I1 = I2 + I3

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Sum of Potential Differences Around a Closed Path

For each element, choice of circulation direction defines

ΔVii∑ = −

Estatic ⋅d

s = 0ClosedPath

∫

ΔV ≡V (after)−V (before) ≡V (a)−V (b)

Sum of potential differences across all elements around any closed circuit loop must be zero.

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RC Charging Circuit

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Charging a Capacitor

What happens when we close switch S at t = 0?

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Group Problem: Charging a Capacitor Using the circuit rules, determine the differential equation satisfied by the charge Q on the positive plate of the capacitor. Your equation should establish a relationship between Q,dQ / dt, R,and ε

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Charging a Capacitor: Solution

dQdt

= − 1̀RC

(Q −Cε )

I = + dQ

dt

ΔVi

i∑ = ε − Q

C− IR = 0

Circulate clockwise

First order linear inhomogeneous differential equation

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Group Problem RC Circuit Charging: Graphical Solution

Given that at time t = 0, Q(t=0) =0, make a qualitative sketch of a plot of

a)  Q(t) vs. t

b)  I(t) vs. (t)

dQdt

= − 1RC

Q − Cε( )

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Solution to Differential Equation for Charging RC Circuits

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Solution to Differential Equation for Charging RC Circuit

dQdt

= − 1RC

Q − Cε( )

dQQ −Cε

= − dtRC

First order linear inhomogeneous differential equation:

Separation of variables:

Integration:

d ′Q′Q −Cε′Q =0

′Q =Q(t )

∫ = − d ′tRCt=0

′t =t

∫ ⇒ ln Q(t)−Cε−Cε

⎛⎝⎜

⎞⎠⎟= − t

RC

Exponentiate:

Q(t)−Cε−Cε

⎛⎝⎜

⎞⎠⎟= e−(t / RC ) ⇒Q(t)−Cε = −Cεe−(t / RC ) ⇒

Q(t) = Cε(1− e−(t / RC ) )

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Time Constant RC Circuit: Charging

Solution to this equation when switch is closed at t = 0 with Q(t = 0) = 0:

τ = RC : time constant (units: seconds)

dQdt

= − 1RC

Q − Cε( )

Q(t) = Cε(1− e− t /τ ) I(t) = + dQ

dt⇒ I(t) = I0e

− t /τ

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Concept Question: Current in RC Circuit

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Concept Question: RC Circuit An uncharged capacitor is connected to a battery, resistor and switch. The switch is initially open but at t = 0 it is closed. A very long time after the switch is closed, the current in the circuit is 1.  Nearly zero 2.  At a maximum and

decreasing 3.  Nearly constant but non-zero

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Concept Q. Answer: RC Circuit

Eventually the capacitor gets “completely charged” – the voltage increase provided by the battery is equal to the voltage drop across the capacitor. The voltage drop across the resistor at this point is 0 – no current is flowing.

Answer: 1. After a long time the current is 0

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Power - Capacitor Circulating across a capacitor from the positive to negative plate decreases your potential. If current flows in that direction the capacitor absorbs power (stores charge)

dtdU

CQ

dtd

CQ

dtdQVIP ===Δ=

2

2

absorbed

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Energy Balance: Circuit Equation

ε − Q

C− IR = 0

Multiplying by

(power delivered by battery) = (power dissipated through resistor) + (power absorbed by the capacitor)

ε I = I 2R + Q

CdQdt

= I 2R + ddt

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Q2

C⎛

⎝⎜⎞

⎠⎟

I = + dQ

dt

I = + dQ

dt

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Concept Question: RC Circuit

Consider the circuit at right, with an initially uncharged capacitor and two identical resistors. At the instant the switch is closed:

1. IR = IC = 02. IR = ε / 2R , IC = 03. IR = 0 , IC = ε / R4. IR = ε / 2R , IC = ε / R

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Concept Question Answer: RC Circuit

Initially there is no charge on the capacitor and hence no voltage drop across it – it looks like a short. Thus all current will flow through it rather than through the bottom resistor. So the circuit looks like:

Answer: 3. IC = ε R IR = 0

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1.  . 2.  3. 

Concept Q.: Current Thru Capacitor In the circuit at right the switch is closed at t = 0. At t = ∞ (long after) the current through the capacitor will be:

IC = 0

IC = ε R

IC = ε 2R

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Con. Q. Ans.: Current Thru Capacitor

After a long time the capacitor becomes “fully charged.” No more current flows into it.

IC = 0Answer 1.

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1.  . 2.  3. 

Concept Q.: Current Thru Resistor

In the circuit at right the switch is closed at t = 0. At t = ∞ (long after) the current through the lower resistor will be:

IR = 0

IR = ε R

IR = ε 2R

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Concept Q. Ans.: Current Thru Resistor

Since the capacitor is “fullly charged” we can remove it from the circuit, and all that is left is the battery and two resistors. So the current is . IR = ε 2R

Answer 3.

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Review Some Math: Exponential Decay

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Math Review: Exponential Decay

dAdt

= − 1τ

AConsider function A where:

A decays exponentially:

A(t) = A0e− t τ

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Exponential Behavior

dAdt

= − 1τ

( A− Af )Slightly modify diff. eq.:

A “grows” to Af: A(t) = Af (1− e− t /τ )

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Time Constant RC Circuit: Charging

Solution to this equation when switch is closed at t = 0:

τ = RC : time constant (units: seconds)

dQdt

= − 1RC

Q − Cε( )

Q(t) = Cε(1− e− t /τ ) I(t) = + dQ

dt⇒ I(t) = I0e

− t /τ

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Demonstration RC Time Constant

Displayed with a Lightbulb (E10) http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%2010&show=0

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RC Discharging Circuit

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Discharging A Capacitor

At t = 0 charge on capacitor is Q0. What happens when we close switch S at t = 0?

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Discharging a Capacitor

ΔVi

i∑ =

QC

− IR = 0

I = −

dQdt

⇒

dQdt

= −QRC

Circulate clockwise

First order linear differential equation

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RC Circuit: Discharging

Solution to this equation when switch is closed at t = 0 with time constant τ = RC

dQdt

= − 1RC

Q ⇒ Q(t) = Qoe− t / RC

I = − dQ

dt⇒ I(t) =

Qo

τe− t /τ =

Qo

RCe− t /τ

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Homework: Solve Differential Equation for Charging and Discharging RC Circuits

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Concept Q.: Open Switch in RC Circuit After the switch has been closed for a very long time, it is opened. What happens to the current through the lower resistor?

1.  It stays the same 2.  Same magnitude, flips direction 3.  It is cut in half, same direction 4.  It is cut in half, flips direction 5.  It doubles, same direction 6.  It doubles, flips direction

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Con. Q. Ans.: Open Switch in RC Circuit

The capacitor has been charged to a potential of , so when it is responsible for pushing current through the lower resistor it pushes a current of , in the same direction as before (its positive terminal is also on the left)

Answer: 1. It stays the same

VC = ε 2

IR = ε 2R

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Group Problem: RC Circuit For the circuit shown in the figure (switch closes at t = 0, opens at t = T>>RC), state the values of currents through the two bottom branches (i) just after switch is closed at t = 0+ (ii) Just before switch is opened at t = T-, (iii) Just after switch is opened at t = T+