[DAY 01 회로이론 기출문제] 오늘은 딱 21문제만 보세요. 더이상 보지 마세요.

Summary

The instructor reviews 21 frequently tested circuit-theory problems. The lesson emphasizes recognizing the circuit or waveform type, choosing the relevant law or formula, and practicing unfamiliar problems before relying on memorized answers.

Recommended study approach

• Scan past-exam questions first to identify recurring problem types.
• If a problem is difficult, try solving it independently two or three times before consulting the worked solution.
• For especially challenging problems, use the textbook’s quick solution as a fallback. If a fixed answer repeatedly appears, memorizing it may help—but understand the method when possible, since CBT versions may change the numbers.
• When reviewing, separate problems you can solve from those you cannot. Hide the solution and test yourself, spending extra time on problems where you repeatedly make mistakes.

Problems and concepts covered

1. Conservation of charge and Kirchhoff’s Current Law (KCL) At a node, the total current entering equals the total current leaving. This expresses conservation of charge, so KCL is the law most closely related to it.

2. A fixed-answer question The instructor presents this as a question whose answer is treated as predetermined. The study tip is to memorize the answer if necessary and move on, while prioritizing understanding for questions that test a reusable method.

3. Mesh-current analysis using Kirchhoff’s Voltage Law (KVL)

   • Assign a current to each loop.
   • Write a KVL equation for each loop, using the difference between mesh currents for resistors shared by two loops.
   • Solve the simultaneous equations.
   • The current through the specified 2 Ω resistor is (I_2), calculated as (28/31) A.

The instructor recommends working through this difficult problem several times before turning to the quick solution.

1. AC bridge in equilibrium Treat the circuit as a Wheatstone-type bridge: at balance, the products of the impedances on opposite arms are equal. Use the inductive and capacitive impedances in that relation; do not multiply component labels such as (L) and (C) without accounting for their impedance forms. The resulting inductance relation is (L=R_1R_2C) for the circuit shown.

2. Equal potentials at points A and B Re-draw the circuit by moving wires while preserving which points are electrically connected. The resulting layout is a Wheatstone bridge. Equal potentials at A and B correspond to the balanced-bridge condition: the products of opposite arms are equal, or equivalently, the arm ratios match.

3. Instantaneous value from a sine-wave graph Read the waveform’s peak value, identify it as a sine wave, and determine its phase from the graph. Then substitute those values into the general form [ v(t)=V_m\sin(\omega t+\theta). ] A wave shifted forward has a positive phase angle.

4. Instantaneous voltage with an initial value and direction

   • Start with (v(t)=V_m\sin(\omega t+\theta)).
   • Use the given value at (t=0) to determine the phase angle.
   • Use whether the waveform is increasing or decreasing to select the correct phase. Here, the increasing condition selects (30^\circ), rather than the other angle with the same sine.
   • Substitute (f=60) Hz and (t=2) ms, converting radians involving (\pi) to degrees consistently.

The resulting angle is (73.2^\circ), so the instantaneous voltage is (10\sin(73.2^\circ)) V.

1. RMS current for a piecewise-constant waveform Use [ I_{\rm rms}=\sqrt{\frac{1}{T}\int_0^T i^2(t)\,dt}. ] For 100 A over 10 seconds followed by 20 A over 20 seconds, add the squared-current areas and divide by the total 30 seconds: [ I_{\rm rms}=\sqrt{\frac{10(100)^2+20(20)^2}{30}}=60\text{ A}. ] The instructor suggests viewing the integral as areas on a graph.

2. Crest factor of a sine wave Crest factor is the peak value divided by the RMS value. For a sine wave, it is (\sqrt{2}), approximately 1.414.

3. Factors for a triangular wave For the triangular waveform discussed:

   - Crest factor: \(\sqrt{3}\approx1.732\).
   - Form factor: \(2/\sqrt{3}\approx1.155\).
   

   These values can be memorized or derived from the peak, RMS, and average values.

4. Identifying an inductor from phase Convert the voltage expression to a sine reference so the phases can be compared. If voltage leads current by (90^\circ)—equivalently, current lags voltage by (90^\circ)—the element is an inductor.

5. RMS current in a pure inductor Find the inductive reactance and then apply (I=V/X_L): [ X_L=2\pi fL,\qquad I=\frac{100}{2\pi(60)(0.1)}\approx2.65\text{ A}. ]

6. Two coils in series Add the resistive and reactive parts: [ Z=(3+j5)+(5+j1)=8+j6\ \Omega. ] With 100 V applied, the current magnitude is (100/|Z|=10) A, with a phase of about (-37^\circ).

7. Power factor in a series R–L circuit Calculate (X_L=2\pi fL), then use [ \cos\theta=\frac{R}{\sqrt{R^2+X_L^2}}. ] With (R=50\,\Omega), (L=0.2) H, and (f=50) Hz, the power factor is about 0.627, or 62.7%.

8. Current in a series R–C circuit Calculate (X_C=1/(2\pi fC)), then use [ I=\frac{V}{\sqrt{R^2+X_C^2}}. ] For the stated values, (X_C\approx88.42\,\Omega), giving a current of about 0.75 A.

9. Power factor in a parallel R–C circuit Use the parallel-circuit expression [ \cos\theta=\frac{1}{\sqrt{1+(\omega CR)^2}}. ] The instructor recommends reviewing the earlier derivation if the formula is unfamiliar.

10. Impedance of a parallel R–C circuit with an exponential voltage Combine the parallel resistor and capacitor impedances to obtain [ Z=\frac{R}{1+sRC}. ] The given exponential voltage determines the value of (s); substituting (s=-5) gives (Z=R/(1-5RC)). The key lesson is to use the condition specified by the problem rather than stop at a general expression.

11. Voltage between two points in an AC circuit Find the complex voltage at A and at B using voltage division, subtract them, and take the magnitude of the result. The worked values are (V_A=32-j24) V and (V_B=18+j24) V, so the difference has magnitude 50 V.

12. Resonance quality factor for a series R–L–C circuit Use [ Q=\frac{1}{R}\sqrt{\frac{L}{C}}. ] Convert milli and micro units carefully before substituting. For the stated values, (Q=10).

13. Equivalent inductance of coupled coils Mutual inductance is [ M=k\sqrt{L_1L_2}. ] For series-aiding coils, (L_{\rm eq}=L_1+L_2+2M); for series-opposing coils, (L_{\rm eq}=L_1+L_2-2M). With two 5 mH coils and (k) ranging up to 0.8, the maximum is 18 mH and the minimum is 2 mH. The same maximum (k) gives the minimum opposing result because it subtracts the largest mutual-inductance term.

14. Identifying aiding or opposing coupled coils from dot markings Use the direction of current relative to the dots to determine the connection: current entering both dotted terminals indicates series aiding; entering a dot on one coil and leaving the dotted terminal of the other indicates series opposing. Choose the corresponding (+2M) or (-2M) formula.

Speakers and sources featured

• One instructor/presenter teaches the circuit-theory review. The channel metadata identifies the presenter as 송건웅 (Song Geon-ung) of 전기는빠지직.
• Hi-Edu is mentioned as the educational brand. The brief boyfriend–girlfriend exchange is an illustrative example spoken by the instructor, not a separate featured speaker.