Chapter 5 Practice Problems

Practice Problems

Problem 1 — Capacitor Combinations

Three capacitors are given: C1 = 10 μF, C2 = 22 μF, C3 = 47 μF.

(a) Calculate the equivalent capacitance when all three are in parallel.

(b) Calculate the equivalent capacitance when all three are in series.

(c) Calculate the equivalent capacitance when C1 is in series with the parallel combination of C2 and C3.

Solution

(a) Parallel capacitors add directly:

\[C_{eq} = 10 + 22 + 47 = 79\text{ μF}\]

(b) Series combination uses the reciprocal formula:

[\frac{1}{C_{eq}} = \frac{1}{10} + \frac{1}{22} + \frac{1}{47} = 0.100 + 0.0455 + 0.0213 = 0.1668\text{ μF}^{-1}] [C_{eq} = \frac{1}{0.1668} = 5.99\text{ μF}]

(c) First find C2 ∥ C3:

\[C_{23} = C_2 + C_3 = 22 + 47 = 69\text{ μF}\]

Then C1 in series with C23:

\[C_{eq} = \frac{C_1 \cdot C_{23}}{C_1 + C_{23}} = \frac{10 \times 69}{10 + 69} = \frac{690}{79} = 8.73\text{ μF}\]

Problem 2 — Capacitor Energy Storage

A 100 μF electrolytic capacitor is charged to 25 V.

(a) How much energy is stored in the capacitor?

(b) The capacitor discharges to 10 V. How much energy was released?

(c) If this energy is released in 5 ms, what is the average power delivered during discharge?

Solution

(a) Energy stored at 25 V:

\[E_{initial} = \frac{1}{2}CV^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (25)^2 = 50 \times 10^{-6} \times 625 = 31.25\text{ mJ}\]

(b) Energy remaining at 10 V:

\[E_{final} = \frac{1}{2} \times 100 \times 10^{-6} \times (10)^2 = 5.0\text{ mJ}\]

Energy released: \(\Delta E = 31.25 - 5.0 = 26.25\text{ mJ}\)

(c) Average power:

\[P = \frac{\Delta E}{\Delta t} = \frac{26.25 \times 10^{-3}}{5 \times 10^{-3}} = 5.25\text{ W}\]

Problem 3 — Inductor Characteristics

An inductor has L = 50 mH. A current that increases linearly from 0 to 2 A in 10 ms is applied.

(a) Calculate the voltage across the inductor during this ramp.

(b) Calculate the energy stored in the inductor at the end of the ramp (when I = 2 A).

(c) If this inductor has a winding resistance of 0.5 Ω, how much power is dissipated in the winding when carrying 2 A DC in steady state?

Solution

(a) The voltage across an inductor is \(v = L\, di/dt\). The rate of change is:

[\frac{di}{dt} = \frac{2\text{ A} - 0}{10 \times 10^{-3}\text{ s}} = 200\text{ A/s}] [v = L \frac{di}{dt} = 50 \times 10^{-3} \times 200 = 10\text{ V}]

(b) Energy stored at 2 A:

\[E_L = \frac{1}{2}LI^2 = \frac{1}{2} \times 50 \times 10^{-3} \times (2)^2 = 0.025 \times 4 = 0.1\text{ J} = 100\text{ mJ}\]

(c) In DC steady state \(di/dt = 0\), so \(v_L = 0\). All voltage drops appear across the winding resistance:

\[P_{winding} = I^2 R_w = (2)^2 \times 0.5 = 2\text{ W}\]

Problem 4 — RMS and Peak Values

A sinusoidal voltage is described by \(v(t) = 170\sin(377t)\) volts.

(a) Identify the peak voltage, angular frequency, frequency, and period.

(b) Calculate the RMS voltage.

(c) This voltage is applied across a 100 Ω resistor. Calculate the average power dissipated.

(d) What is the peak-to-peak voltage of this signal?

Solution

(a) From the equation:

Peak voltage: \(V_m = 170\text{ V}\)
Angular frequency: \(\omega = 377\text{ rad/s}\)
Frequency: \(f = \omega / (2\pi) = 377 / 6.283 = 60\text{ Hz}\) (standard US AC frequency)
Period: \(T = 1/f = 1/60 = 16.7\text{ ms}\)

(b) RMS voltage:

\[V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{170}{\sqrt{2}} = \frac{170}{1.414} = 120.2\text{ V} \approx 120\text{ V}\]

(c) Average power:

\[P = \frac{V_{rms}^2}{R} = \frac{(120)^2}{100} = \frac{14{,}400}{100} = 144\text{ W}\]

(d) Peak-to-peak voltage:

\[V_{pp} = 2V_m = 2 \times 170 = 340\text{ V}\]

Problem 5 — Decibel Calculations

An audio amplifier has an input signal of 10 mV and produces an output of 5 V.

(a) Calculate the voltage gain in decibels.

(b) A signal passes through two cascaded stages: Stage 1 has a gain of +20 dB, Stage 2 has a gain of −6 dB. What is the overall voltage gain in dB? What is the overall voltage ratio?

(c) A power amplifier increases signal power from 1 mW to 50 W. Express this as a power gain in decibels.

Solution

(a) Voltage gain in dB:

\[A_v(\text{dB}) = 20\log_{10}\left(\frac{V_{out}}{V_{in}}\right) = 20\log_{10}\left(\frac{5}{0.010}\right) = 20\log_{10}(500) = 20 \times 2.699 = 53.98\text{ dB}\]

(b) Cascaded gains add in dB:

\[A_{total} = +20 + (-6) = +14\text{ dB}\]

Converting back to voltage ratio: \(V_{ratio} = 10^{14/20} = 10^{0.7} = 5.01\)

(c) Power gain in dB:

\[G_P = 10\log_{10}\left(\frac{P_{out}}{P_{in}}\right) = 10\log_{10}\left(\frac{50}{0.001}\right) = 10\log_{10}(50{,}000) = 10 \times 4.699 = 46.99\text{ dB} \approx 47\text{ dB}\]

Problem 6 — Mutual Inductance

Two coupled inductors have L1 = 100 mH, L2 = 400 mH, and a coupling coefficient k = 0.5.

(a) Calculate the mutual inductance M.

(b) If the current in L1 changes at a rate of 500 A/s, what voltage is induced in L2?

(c) If both inductors are connected in series with aiding flux (currents in same direction), what is the total inductance?

Solution

(a) Mutual inductance:

\[M = k\sqrt{L_1 L_2} = 0.5\sqrt{0.1 \times 0.4} = 0.5\sqrt{0.04} = 0.5 \times 0.2 = 0.1\text{ H} = 100\text{ mH}\]

(b) Induced voltage in L2:

\[v_{12} = M\frac{di_1}{dt} = 0.1 \times 500 = 50\text{ V}\]

(c) Series-aiding inductors:

\[L_{total} = L_1 + L_2 + 2M = 100 + 400 + 2(100) = 700\text{ mH}\]