Rc circuits
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RC Circuits. AP Physics C Montwood High School R. Casao. RC Circuits. To date, we have studied steady-state direct current circuits in which the current is constant. In circuits containing a capacitor, the current varies over time.

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Rc circuits

RC Circuits

AP Physics C

Montwood High School

R. Casao


Rc circuits1

RC Circuits

  • To date, we have studied steady-state direct current circuits in which the current is constant.

  • In circuits containing a capacitor, the current varies over time.

  • When a potential difference is applied across a capacitor, the rate at which the capacitor charges depends on the capacitance and on the resistance in the circuit.

  • An RC circuit can store charge and release it at a later time.


Charging a capacitor

Charging a Capacitor

  • Consider a series circuit containing a resistor and a capacitor that is initially uncharged.

  • With switch S open, there is no current in the circuit.

  • When switch S is closed at t = 0 s, charges begin to flow and a current is present in the circuit and the capacitor begins to charge.


Charging a capacitor1

Charging a Capacitor

  • The gap between the capacitor plates represents an open circuit and charge does not pass from the positive plate to the negative plate.

  • Charge is transferred from one plate to the other plate through the resistor, switch, and battery until the capacitor is fully charged.

  • The value of the maximum charge depends on the EMF of the battery.

  • Once the maximum charge is reached, the current in the circuit is zero.


Charging a capacitor2

Charging a Capacitor

  • Applying Kirchhoff’s loop rule to the circuit after the switch is closed:

  • I ·R is the potential drop across the resistor.

  • q/C is the potential drop across the capacitor.

  • I and q are instantaneous values of the current and charge as the capacitor charges.


Charging a capacitor3

Charging a Capacitor

  • At t = 0 s, when the switch is closed, the charge on the capacitor is 0 C and the initial current is:

  • At t = 0 s, the potential drop is entirely across the resistor.

  • As the capacitor is charged to its maximum value Q, the charges quit flowing and the current in the circuit is 0 A and the potential drop is entirely across the capacitor.


Charging a capacitor4

Charging a Capacitor

  • Maximum charge:

  • From t = 0 s until the capacitor is fully charged and the current stops, the amount of current in the circuit decreases over time and the amount of charge on the capacitor increases over time.

  • To determine values for the current in the circuit and for the charge on the capacitor as functions of time, we have to use a differential equation.


Charging a capacitor current equation

Charging a Capacitor – Current Equation

  • Beginning with Kirchhoff’s loop equation, differentiate the equation with respect to time:


Charging a capacitor current equation1

Charging a Capacitor – Current Equation

  • Replace dq/dt with I:

  • Get the current terms on one side of the equation and the other terms on the other side of the equation:

  • Integrate both sides of the equation.


Charging a capacitor current equation2

Charging a Capacitor – Current Equation

  • The limits of integration for the current side of the equation is from Imax (at t = 0 s) to the current value at time t.

  • The limits of integration for the time side of the equation is from t = 0 s to time t.


Charging a capacitor current equation3

Charging a Capacitor – Current Equation

  • Left side:

  • Right side:


Charging a capacitor current equation4

Charging a Capacitor – Current Equation

  • Combining both sides of the integration:

  • To eliminate the natural log term (ln), we can use the terms as exponents for the base e. From the properties of logarithms:


Charging a capacitor current equation5

Charging a Capacitor – Current Equation

  • Current in an RC circuit as a function of time:

  • Graph of Current vs. time for a charging capacitor:


Charging a capacitor charge equation

Charging a Capacitor – Charge Equation

  • To find the charge on the capacitor as a function of time, begin with Kirchhoff’s loop equation:

  • Replace I with dq/dt:


Charging a capacitor charge equation1

Charging a Capacitor – Charge Equation

  • Get the dq/dt term on one side of the equation:

  • Divide both sides by R to solve for dq/dt:

  • Common demominator, R·C:


Charging a capacitor charge equation2

Charging a Capacitor – Charge Equation

  • Multiply both sides by dt:

  • Divide both sides by E·C-q to get the charge terms together on the same side of the equation:

  • In problems involving capacitance, q is positive, so multiply both sides by –1 to make the charge positive:


Charging a capacitor charge equation3

Charging a Capacitor – Charge Equation

  • Integrate both sides of the resulting differential equation.

    • For the charge side of the equation, the limits of integration are from q = 0 at t = 0 s to q at time t. I rearranged the equation to put the positive q first followed by the negative E·C.

    • For the time side of the equation, the limits of integration are from 0 s to t.


Charging a capacitor charge equation4

Charging a Capacitor – Charge Equation

  • Left side:

  • Right side:


Charging a capacitor charge equation5

Charging a Capacitor – Charge Equation

  • Combining both sides of the integrals:

  • To eliminate the natural log term (ln), we can use the terms as exponents for the base e.


Charging a capacitor charge equation6

Charging a Capacitor – Charge Equation

  • Simplify:

  • Solve for q:

    • Multiply both sides by –E·C.

    • Add E·C to both sides.

  • Factor out the E·C:


Charging a capacitor charge equation7

Charging a Capacitor – Charge Equation

  • Substitute: Qmax = E·C

  • Graph of Charge vs. time for a charging capacitor:


Charging a capacitor5

Current has its maximum value of I max = E/R at t = 0 s and decays exponentially to 0 A as t infinity.

Charging A Capacitor


Charging a capacitor6

The charge on a capacitor is 0 C at t = 0 s and approaches a maximum value of Qmax = C·E as t infinity.

Charging A Capacitor


Rc circuits

Charging Capacitor Graphs

current voltage charge


Time constant r c

Time Constant R·C

  • The quantity R·C, which appears in the exponential component of the charge and current equations is called the time constant t of the circuit.

  • The time constant is a measure of how quickly the capacitor becomes charged.

  • The time constant represents the time it takes the:

    • current to decrease to 1/e of its initial value.

    • charge to increase from 0 C to C·E·(1-e-1) = 0.63·C·E.

  • The unit for the time constant is seconds.

    Ω · F = (V/A)(C/V) = C/(C/s) = s


  • Charge and current during the charging of a capacitor

    q

    Qmax

    Capacitor

    Capacitor

    i

    Rise in Charge

    I

    0.37 I

    Current Decay

    Time, t

    t

    0.63 I

    Time, t

    t

    Charge and Current during the Charging of a Capacitor

    In a time t of one time constant, the chargeq rises to63%of its maximum, while the currenti decays to 37%of its maximum value.


    Discharging a capacitor

    Discharging a Capacitor

    • Removing the battery from the circuit while keeping the switch open leaves us with a circuit containing only a charged capacitor and a resistor.


    Discharging a capacitor1

    Discharging a Capacitor

    • When the switch is open, there is a potential difference of Q/C across the capacitor and 0 V across the resistor since I = 0 A.

    • If the switch is closed at time t = 0 s, the capacitor begins to discharge through the resistor and a current flows through the circuit.

    • At some time during the discharge, current in the circuit is I and the charge on the capacitor is q.


    Discharging a capacitor charge equation

    Discharging a Capacitor – Charge Equation

    • To find the charge on the capacitor as a function of time, begin with Kirchhoff’s loop equation. There is no E term in the equation because the battery has been removed. The I·R term is negative because the energy carried by the charges is dissipated in the resistor.


    Discharging a capacitor charge equation1

    Discharging a Capacitor – Charge Equation

    • Replace I with -dq/dt because the current in the circuit is decreasing as the capacitor discharges over time:

    • Get the charge terms on one side of the equation and the remaining variables on the other side of the equation.


    Discharging a capacitor charge equation2

    Discharging a Capacitor – Charge Equation

    • Integrate both sides of the resulting differential equation.

      • For the charge side of the equation, the limits of integration are from q = Qmax at t = 0 s to q at time t.

      • For the time side of the equation, the limits of integration are from 0 s to t.


    Discharging a capacitor charge equation3

    Discharging a Capacitor – Charge Equation

    • Left side:

    • Right side:


    Discharging a capacitor charge equation4

    Discharging a Capacitor – Charge Equation

    • Combining both sides of the integrals:

    • To eliminate the natural log term (ln), we can use the terms as exponents for the base e.


    Discharging a capacitor charge equation5

    Discharging a Capacitor – Charge Equation

    • Simplify:

    • Solve for q:


    Discharging a capacitor current equation

    Discharging a Capacitor – Current Equation

    • To find the current on the capacitor as a function of time, begin with the charge equation.

    • Current I = dq/dt, so take the derivative of the charge equation with respect to time:


    Discharging a capacitor current equation1

    Discharging a Capacitor – Current Equation

    • Left side:

    • Right side:


    Discharging a capacitor current equation2

    Discharging a Capacitor – Current Equation

    • Combining the two sides of the integral:

    • Qmax = C·V, substituting:


    Discharging a capacitor current equation3

    Discharging a Capacitor – Current Equation

    • The negative sign indicates that the direction of the discharging current is opposite to the direction of the charging current.

    • The voltage V across the capacitor is equal to the EMF of the battery since the capacitor is fully charged at the time of the switch is closed to discharge the capacitor through the resistor.


    Rc circuits

    Discharging Capacitor Graphs

    voltage charge current


    Bonus equations

    Bonus Equations!

    • I have never seen these equations in any textbook and had never been asked to find the voltage across the capacitor as a function of time. I got these equations from the E & M course I took Fall 06.

    • Discharging Capacitor:


    Bonus equations1

    Bonus Equations!

    • Charging Capacitor:


    Energy conservation in charging a capacitor

    Energy Conservation in Charging a Capacitor

    • During the charging process, a total charge Q = EC flows thru the battery.

    • The battery does work W = QmaxE or W = CE2.

    • Half of this work is accounted for by the energy stored in the capacitor:

      U = 0.5QV = 0.5QmaxE = 0.5CE2

    • The other half of the work done by the battery goes into Joule heat in the resistance of the circuit.


    Rc circuits

    • The rate at which energy is put into the resistance R is:

    • Using the equation for current in a charging capacitor:

    • Determine the total Joule heat by integrating from t = 0 s to t = :


    Rc circuits

    • Substitute: let


    Rc circuits

    • The result is independent of the resistance R; when a capacitor is charged by a battery with a constant EMF, half the energy provided by the battery is stored in the capacitor and half goes into thermal energy.

    • The thermal energy includes the energy that goes into the internal resistance of the battery.


    Capacitors and resistors in parallel

    Capacitors and Resistors in Parallel

    • The capacitor in the figure is initially uncharged when the switch S is closed.

    • Immediately after the switch is closed, the potential is the same at points c and d.


    Rc circuits

    • An uncharged capacitor does not resist the flow of current and acts like a wire.

    • No current flows thru the 8 Ωresistor; the

      capacitor acts as a short circuit between points c and d.

    • Apply Kirchhoff’s loop rule to the outer loop abcdefa: 12 V – 4 Ω·I0 = 0; I0 = 3 A


    Rc circuits

    • After the capacitor is fully charged, no more charge flows onto or off of the plates; the capacitor acts like a broken wire or open in the circuit.

    • Apply Kirchhoff’s loop rule to loop abefa:

      12 V – 4 Ω·If– 8 Ω·If = 0; 12 V – 12 Ω·If; If = 1 A


    Rc circit problem example

    RC Circit Problem Example

    • You will see this problem on your homework.


    Rc circuits

    • Charging: When the switch S is first closed and the current begins to move through the circuit, it will move through the 2 Ω resistor onto the capacitor.

    • Charge will move off of the capacitor through the 22 Ω to the battery.

    • No charge moves through the 18 Ω or 38 Ω resistor while the capacitor charges.


    Rc circuits

    • Once the voltage across the 22 μF capacitor reaches its maximum voltage, no additional charge will move onto the capacitor and the current in the circuit will now begin to move through all four resistors.

    • The 18 Ω and 22 Ω resistors are in series; total resistance = 40 Ω .

    • The 2 Ω and 38 Ω resistors are in series; total resistance = 40 Ω.


    Rc circuits

    • The 40 Ω resistors are in parallel with each other; total resistance of the circuit:

    • Total current:

    • The current in each branch is:


    Rc circuits

    • The voltage drop across the 18 Ω resistor is

      V = 0.5 A·18 Ω = 9 V.

    • The voltage drop across the 2 Ω resistor is V = 0.5 A·2 Ω = 1 V.

    • The difference in

      the voltage

      between the two

      points is equal to

      the voltage across

      the capacitor: 9V – 1 V = 8 V.


    Rc circuits

    • Check: determine the voltage difference between the 22 Ω and the 38 Ω resistor:

    • ΔV = 19 V – 11 V = 8 V

    • The difference in the voltage between these two resistors is equal to the voltage across the capacitor.


    Rc circuits

    • Discharging: when the switch S is opened and the battery is removed from the circuit, the capacitor will discharge current I1 through the 18 Ω and 2 Ω resistors and current I2 through the 22 Ω and 38 Ω resistors.

    • The 18 Ω and 2 Ω resistors are in series; total resistance = 20 Ω.

    • The 22 Ω and 38 Ω resistors are in series; total resistance = 60 Ω.

    • The 20 Ω and and 60 Ω resistances are in parallel with each other.


    Rc circuits

    • Total resistance:

    • Time constant τ = R·C = 15 Ω·22 x 10-6 F = 3.3 x 10-4 s

    • From here, you can use the discharging voltage equation to solve for the time t.


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