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Omniguides by NOX
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Chapter 4 Solutions

These worked solutions are authored for the reviewed prompts shown here. Any added geometry or assumption is stated. Original chapter problems supplied no answer key, so these solutions are separate from the archived manuscript.

Original problem 1 · worked solution

An ideal 12 V source is connected across C₁=2 μF in series with a parallel pair C₂=C₃=3 μF. All capacitors start uncharged. Find the equivalent capacitance.

Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

  1. Combine the parallel pair.

    C23=C2+C3=6 μFC_{23}=C_2+C_3=6\,\mu\mathrm F.

  2. Combine C₁ and C₂₃ in series.

    Ceq=(1/2+1/6)−1=1.50 μFC_{\rm eq}=(1/2+1/6)^{-1}=1.50\,\mu\mathrm F.
Answer

Ceq=1.50 μF.

Check

A series equivalent is smaller than either series member, 2 μF and 6 μF.

Practice prompt · Original reading location

Archived original prompt
  • Determine the total capacitance of the network in blah.

    • Original problem 2 · worked solution

    An ideal 12 V source is connected across C₁=2 μF in series with a parallel pair C₂=C₃=3 μF. All capacitors start uncharged. At electrostatic equilibrium, find the voltage across each capacitor.

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    1. Use the equivalent capacitance from the stated network.

      Q=CeqV=(1.50 μF)(12 V)=18 μCQ=C_{\rm eq}V=(1.50\,\mu\mathrm F)(12\,\mathrm V)=18\,\mu\mathrm C.

    2. Series members carry the same terminal charge magnitude.

      V1=Q/C1=9.0 V,V23=Q/C23=3.0 VV_1=Q/C_1=9.0\,\mathrm V,\quad V_{23}=Q/C_{23}=3.0\,\mathrm V.

    3. Parallel members share voltage.

      V₂=V₃=V₂₃=3.0 V.
    Answer

    V₁=9 V, V₂=3 V and V₃=3 V.

    Check

    The series drops add to 12 V. Parallel charges Q₂=Q₃=9 μC add to the 18 μC group charge.

    Practice prompt · Original reading location

    Archived original prompt
  • Determine the voltages across each capacitor shown in the circuit in blah using the given information and the capacitance calculated in the previous problem.

    • Original problem 3 · worked solution

    An ideal 12 V source drives R₁=4 Ω in series with a parallel pair R₂=6 Ω and R₃=3 Ω. Find the equivalent resistance.

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    1. Combine the parallel resistors.

      R23=(1/6+1/3)−1=2 ΩR_{23}=(1/6+1/3)^{-1}=2\,\Omega.

    2. Add the series member.

      Req=R1+R23=6 ΩR_{\rm eq}=R_1+R_{23}=6\,\Omega.
    Answer

    Req=6 Ω.

    Check

    The parallel equivalent 2 Ω is smaller than either branch resistor.

    Practice prompt · Original reading location

    Archived original prompt
  • Determine the total resistance of the circuit depicted in blah

    • Original problem 4 · worked solution

    An ideal 12 V source drives R₁=4 Ω in series with a parallel pair R₂=6 Ω and R₃=3 Ω. Find each resistor current.

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    1. The series current follows from total resistance.

      I1=12/6=2 AI_1=12/6=2\,\mathrm A.

    2. Find the parallel-group voltage.

      V23=I1R23=4 VV_{23}=I_1R_{23}=4\,\mathrm V.

    3. Use Ohm’s law on each branch.

      I2=4/6=2/3 A,I3=4/3 AI_2=4/6=2/3\,\mathrm A,\quad I_3=4/3\,\mathrm A.
    Answer

    I₁=2 A, I₂=2/3 A and I₃=4/3 A.

    Check

    KCL gives I₂+I₃=2 A, and KVL gives the series drops 8 V+4 V=12 V.

    Practice prompt · Original reading location

    Archived original prompt
  • Determine the currents across each of the resistors depicted in the circuit in blah

    • Original problem 5 · worked solution

    Optional RC supplement: arbitrary-time exponential formulas. Prerequisites: exponentials and the time constant τ=RC; the optional calculus derivation also uses logarithms and derivatives. This item makes no AP coverage claim.

    An ideal single-loop RC circuit has R=2.00 kΩ and C=100 μF, so τ=RC=0.200 s. At t=0 an uncharged capacitor is connected to a 12 V ideal source. Find q(t).

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    The arbitrary-time charging formula can be used as a supplied supplement result. The separation-of-variables steps below are an optional calculus derivation; the initial condition fixes the integration constant.

    1. Write the charging loop equation.

      V0−R dq/dt−q/C=0,V0=12 VV_0-R\,dq/dt-q/C=0,\qquad V_0=12\,\mathrm V.

    2. Express the rate in terms of the final charge and time constant.

      Let Qf=CV0,τ=RCQ_f=CV_0,\quad\tau=RC. Multiplying the loop equation by C and dividing by RC gives dqdt=Qf−qτ{dq\over dt}={Q_f-q\over\tau}.

    3. For the optional calculus derivation, separate variables and include the initial condition.

      For finite t≥0, q starts at 0 and remains below Q_f. With u and s as integration variables, ∫0q(t)duQf−u=∫0tdsτ\int_0^{q(t)}{du\over Q_f-u}=\int_0^t{ds\over\tau}.

    4. Evaluate the integrals as a dimensionless logarithm.

      −ln⁡ ⁣(Qf−q(t)Qf)=tτ-\ln\!\left({Q_f-q(t)\over Q_f}\right)={t\over\tau}. The ratio inside the logarithm is positive and unitless.

    5. Exponentiate and isolate the charge.

      Qf−q(t)Qf=e−t/τ,q(t)=Qf(1−e−t/τ){Q_f-q(t)\over Q_f}=e^{-t/\tau},\qquad q(t)=Q_f(1-e^{-t/\tau}).

    6. Insert the given capacitance and time constant.

      q(t)=1.20 mC[1−e−t/(0.200 s)]q(t)=1.20\,\mathrm{mC}[1-e^{-t/(0.200\,\mathrm s)}].
    Answer

    q(t)=1.20 mC[1−e−t/(0.200 s)]q(t)=1.20\,\mathrm{mC}[1-e^{-t/(0.200\,\mathrm s)}].

    Check

    q(0)=0 and q(∞)=1.20 mC. This single-loop DC transient retains one current direction.

    Practice prompt · Original reading location

    Archived original prompt
  • Find an equation which describes the charging of circuit blah using the appropriate resistances and capacitances

    • Original problem 6 · worked solution

    Optional RC supplement: arbitrary-time exponential formulas. Prerequisites: exponentials and the time constant τ=RC; the optional calculus derivation also uses logarithms and derivatives. This item makes no AP coverage claim.

    An ideal single-loop RC circuit has R=2.00 kΩ and C=100 μF, so τ=RC=0.200 s. A capacitor initially at 12 V is disconnected from its source and discharged through R at t=0. Find q(t).

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    The arbitrary-time discharge formula can be used as a supplied supplement result. The logarithm and exponential steps below are an optional calculus derivation.

    1. The initial charge is fixed by the initial voltage.

      Q0=CV0=1.20 mCQ_0=CV_0=1.20\,\mathrm{mC}.

    2. Charge leaves the positive plate through the resistor.

      dqdt=−qτ,τ=RC{dq\over dt}=-{q\over\tau},\qquad\tau=RC.

    3. For the optional calculus derivation, separate variables using q(0)=Q₀.

      For finite t≥0 the positive charge permits division by q: ∫Q0q(t)duu=−∫0tdsτ\int_{Q_0}^{q(t)}{du\over u}=-\int_0^t{ds\over\tau}. Therefore ln⁡ ⁣(q(t)Q0)=−tτ\ln\!\left({q(t)\over Q_0}\right)=-{t\over\tau}.

    4. Exponentiate and insert the given time constant.

      q(t)=Q0e−t/τ=1.20 mCe−t/(0.200 s)q(t)=Q_0e^{-t/\tau}=1.20\,\mathrm{mC}e^{-t/(0.200\,\mathrm s)}.
    Answer

    q(t)=1.20 mCe−t/(0.200 s)q(t)=1.20\,\mathrm{mC}e^{-t/(0.200\,\mathrm s)}.

    Check

    The charge decreases monotonically to zero; energy is converted to Joule heat.

    Practice prompt · Original reading location

    Archived original prompt
  • Find an equation which describes the discharge of circuit blah using the appropriate resistances and capacitances

    • Original problem 7 · worked solution

    Optional RC supplement: arbitrary-time exponential formulas. Prerequisites: exponentials and the time constant τ=RC; the optional calculus derivation also uses logarithms and derivatives. This item makes no AP coverage claim.

    An ideal single-loop RC circuit has R=2.00 kΩ and C=100 μF, so τ=RC=0.200 s. The initially uncharged capacitor begins charging from a 12 V ideal source. Find the source current I(t), positive toward the capacitor’s positive plate.

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    1. Differentiate the charging charge.

      I(t)=dq/dt=(12/R)e−t/(RC)I(t)=dq/dt=(12/R)e^{-t/(RC)}.

    2. Use R=2000 Ω and RC=0.200 s.

      I(t)=6.00 mAe−t/(0.200 s)I(t)=6.00\,\mathrm{mA}e^{-t/(0.200\,\mathrm s)}.
    Answer

    I(t)=6.00 mAe−t/(0.200 s)I(t)=6.00\,\mathrm{mA}e^{-t/(0.200\,\mathrm s)}.

    Check

    I(0)=V/R and I(∞)=0, while the capacitor charge rises.

    Practice prompt · Original reading location

    Archived original prompt
  • Determine the total current going through circuit blah while the capacitors are charging

    • Original problem 8 · worked solution

    Optional RC supplement: arbitrary-time exponential formulas. Prerequisites: exponentials and the time constant τ=RC; the optional calculus derivation also uses logarithms and derivatives. This item makes no AP coverage claim.

    An ideal single-loop RC circuit has R=2.00 kΩ and C=100 μF, so τ=RC=0.200 s. The source is removed and a capacitor initially at 12 V discharges through R. Find the current magnitude I_out(t) flowing out of its positive plate as a function of time for t≥0, with time measured from source removal at t=0.

    Authored clarification: The original refers to “blah” without a recoverable named circuit. This is an explicitly authored replacement with a complete stated topology and values; no original diagram or source answer is inferred.

    1. Current out of the positive plate is the rate of charge loss.

      Iout(t)=−dq/dt=Q0e−t/(RC)/(RC)I_{\rm out}(t)=-dq/dt=Q_0e^{-t/(RC)}/(RC).

    2. Use Q₀=C(12 V).

      Iout(t)=(12/R)e−t/(RC)=6.00 mAe−t/(0.200 s)I_{\rm out}(t)=(12/R)e^{-t/(RC)}=6.00\,\mathrm{mA}e^{-t/(0.200\,\mathrm s)}.
    Answer

    Iout(t)=6.00 mAe−t/(0.200 s)I_{\rm out}(t)=6.00\,\mathrm{mA}e^{-t/(0.200\,\mathrm s)}.

    Check

    Both charge and current magnitude decrease; if current were defined toward the positive plate, its signed value would be negative.

    Practice prompt · Original reading location

    Archived original prompt
  • Determine the total current going through circuit blah while the capacitors are discharging
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