Chapter 5 Practice Problems — Passive Components
These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.
Problem 1 — Series and Parallel Capacitors
Three capacitors have values \(C_1 = 10\) μF, \(C_2 = 22\) μF, and \(C_3 = 47\) μF.
(a) Find the equivalent capacitance when all three are connected in parallel.
(b) Find the equivalent capacitance when all three are connected in series.
(c) A 12 V source is connected across the parallel combination. How much total charge is stored, and how much energy?
Hint
(a) Capacitors in parallel add directly (like resistors in series): \(C_{eq} = C_1 + C_2 + C_3\). The parallel combination has more plate area, so capacitance increases.
(b) Series capacitors use the reciprocal formula: \(1/C_{eq} = 1/C_1 + 1/C_2 + 1/C_3\). The result will be smaller than the smallest individual capacitor.
(c) Total charge: \(Q = C_{eq,parallel} \cdot V\). Energy stored: \(W = \frac{1}{2} C_{eq,parallel} V^2\). Remember to convert μF to F before computing energy in joules.
Problem 2 — Series and Parallel Inductors
Two inductors, \(L_1 = 15\) mH and \(L_2 = 60\) mH, are connected with no mutual coupling.
(a) Find the equivalent inductance when they are in series.
(b) Find the equivalent inductance when they are in parallel.
(c) At DC steady state, a 6 A current flows through the series combination. Calculate the energy stored in each inductor and in the series combination.
Hint
(a) Inductors in series add like resistors: \(L_{eq} = L_1 + L_2\). With no mutual coupling, this is straightforward.
(b) Inductors in parallel use the reciprocal formula: \(L_{eq} = \frac{L_1 L_2}{L_1 + L_2}\).
(c) Energy in each: \(W_k = \frac{1}{2} L_k I^2\). Since the same 6 A flows through both (series), compute each separately. The total should equal \(\frac{1}{2} L_{total} I^2\) as a check. Convert mH to H.
Problem 3 — Energy Stored and the Capacitor V-I Relationship
A 470 μF capacitor is initially uncharged. Starting at t = 0, a constant current of 2 mA is forced through it.
(a) Write the expression for \(V_C(t)\) using the integral form of the capacitor V-I relationship.
(b) How long does it take for the capacitor to reach 10 V?
(c) How much energy is stored in the capacitor at that moment?
Hint
(a) The integral form is \(v_C(t) = \frac{1}{C}\int_0^t i \, d\tau + v_C(0)\). For a constant current \(i = I\), this simplifies to \(v_C(t) = \frac{I}{C} t\) (a linear ramp from zero).
(b) Set \(v_C(t) = 10\) V and solve for t: \(t = \frac{10 \cdot C}{I}\). Be careful with units: C in farads, I in amperes.
(c) Use \(W = \frac{1}{2} C V^2\) with \(V = 10\) V. Express the result in millijoules.
Problem 4 — Real Inductor Model and RC Time Constant
A real inductor is modeled as an ideal 50 mH inductor in series with its winding resistance \(R_{DCR} = 8\) Ω.
(a) This real inductor is connected in series with an external resistor \(R_{ext} = 42\) Ω and a 12 V DC source. Find the DC steady-state current.
(b) What is the time constant \(\tau\) of this RL circuit? (Use the total series resistance.)
(c) At DC steady state, how much energy is stored in the inductor's magnetic field? How does the winding resistance affect this compared to an ideal inductor with no DCR?
Hint
(a) At DC steady state, the inductor is a short circuit (zero voltage drop). The only opposition is the total series resistance: \(R_{total} = R_{DCR} + R_{ext}\). Apply Ohm's Law.
(b) \(\tau = L / R_{total}\). Both the DCR and the external resistor contribute to the total resistance the circuit "sees" when computing the time constant.
(c) Energy: \(W = \frac{1}{2} L I^2\). The winding resistance DCR reduces the steady-state current compared to an ideal inductor (which would draw more current with only \(R_{ext}\)), so it also reduces the stored energy.