201. Consider the following statements about pure \(R\), \(L\), and \(C\) circuits.
I. In pure \(R\), voltage and current are in phase.
II. In pure \(L\), current leads voltage by \(\frac{\pi}{2}\).
III. In pure \(C\), current leads voltage by \(\frac{\pi}{2}\).
IV. In pure \(L\) and pure \(C\), ideal average power is zero.
ⓐ. I, II, and IV only
ⓑ. II and III only
ⓒ. I, II, III, and IV
ⓓ. I, III, IV only
Correct Answer: I, III, IV only
Explanation: Statement I is true because a pure resistor has \(v\) and \(i\) in phase. Statement II is false because in a pure inductor, current lags voltage by \(\frac{\pi}{2}\), not leads. Statement III is true because in a pure capacitor, current leads voltage by \(\frac{\pi}{2}\). Statement IV is true for ideal pure reactive elements because they store and return energy without net dissipation. The lead-lag direction is the main point that separates the inductive and capacitive cases.
202. A series circuit is formed by connecting \(R\), \(L\), and \(C\) one after another to a sinusoidal \(\text{AC}\) source. The current through the three elements is
ⓐ. the same at every instant in all three elements
ⓑ. different in all three elements because their voltage drops differ
ⓒ. maximum in the capacitor only
ⓓ. zero in the inductor only
Correct Answer: the same at every instant in all three elements
Explanation: In a series circuit, there is only one path for charge flow. Therefore, the same current passes through \(R\), \(L\), and \(C\). The voltage drops across the three elements can have different magnitudes and phases, but the current is common. This common-current feature is why series \(LCR\) phasor diagrams often take current as the reference. Confusing different voltage drops with different series currents gives an incorrect circuit picture.
203. The source voltage in a series \(LCR\) circuit is obtained from the component voltages by
ⓐ. ordinary arithmetic addition of \(V_R\), \(V_L\), and \(V_C\) in all cases
ⓑ. phasor addition of \(V_R\), \(V_L\), and \(V_C\)
ⓒ. multiplying \(V_R\), \(V_L\), and \(V_C\)
ⓓ. taking only the largest of \(V_R\), \(V_L\), and \(V_C\)
Correct Answer: phasor addition of \(V_R\), \(V_L\), and \(V_C\)
Explanation: In a series \(LCR\) circuit, the current is common but the voltage drops across \(R\), \(L\), and \(C\) are not all in the same phase. \(V_R\) is in phase with current, \(V_L\) leads current by \(\frac{\pi}{2}\), and \(V_C\) lags current by \(\frac{\pi}{2}\). Because of these phase differences, the source voltage must be found by phasor addition. Simple arithmetic addition would ignore direction in the phasor diagram. The total voltage is a vector-like resultant in phase space.
204. With current chosen as the reference phasor in a series \(LCR\) circuit, the voltage across the resistor \(V_R\) is drawn
ⓐ. \(90^\circ\) ahead of current
ⓑ. \(90^\circ\) behind current
ⓒ. opposite to current by \(180^\circ\)
ⓓ. along the current phasor
Correct Answer: along the current phasor
Explanation: The resistor voltage is related to current by \(V_R=IR\). In a resistor, voltage and current are in phase. Therefore, when current is chosen as the reference phasor, \(V_R\) is drawn along the same direction. The inductor voltage \(V_L\) is drawn \(90^\circ\) ahead of current, while the capacitor voltage \(V_C\) is drawn \(90^\circ\) behind current. This reference choice makes the series \(LCR\) phasor diagram easier to construct.
205. Using current as reference in a series \(LCR\) circuit, \(V_L\) and \(V_C\) are
ⓐ. along the same direction as \(V_R\)
ⓑ. always equal in magnitude at every frequency
ⓒ. opposite phasors along the reactive axis
ⓓ. both opposite to current by \(180^\circ\)
Correct Answer: opposite phasors along the reactive axis
Explanation: The inductor voltage \(V_L\) leads current by \(\frac{\pi}{2}\), while the capacitor voltage \(V_C\) lags current by \(\frac{\pi}{2}\). With current as reference, these two voltages lie on the same reactive line but in opposite directions. Their magnitudes need not be equal except under the resonance condition. The resistor voltage \(V_R\) lies along the current direction, not along the reactive axis. The opposition of \(V_L\) and \(V_C\) is the basis of net reactive voltage \(V_L-V_C\).
206. For a series \(LCR\) circuit with \(I_{\text{rms}}=2\,\text{A}\), \(R=30\,\Omega\), \(X_L=50\,\Omega\), and \(X_C=20\,\Omega\), the rms voltage drops \(V_R\), \(V_L\), and \(V_C\) are respectively
ⓐ. \(60\,\text{V}\), \(40\,\text{V}\), \(100\,\text{V}\)
ⓑ. \(30\,\text{V}\), \(50\,\text{V}\), \(20\,\text{V}\)
ⓒ. \(100\,\text{V}\), \(60\,\text{V}\), \(40\,\text{V}\)
ⓓ. \(60\,\text{V}\), \(100\,\text{V}\), \(40\,\text{V}\)
Correct Answer: \(60\,\text{V}\), \(100\,\text{V}\), \(40\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(I_{\text{rms}}=2\,\text{A}\), \(R=30\,\Omega\), \(X_L=50\,\Omega\), and \(X_C=20\,\Omega\).
\( \textbf{Required:} \) \(V_R\), \(V_L\), and \(V_C\).
\( \textbf{Voltage across resistor:} \)
\[
V_R=I_{\text{rms}}R
\]
\( \textbf{Substitution for \(V_R\):} \)
\[
V_R=(2)(30)=60\,\text{V}
\]
\( \textbf{Voltage across inductor:} \)
\[
V_L=I_{\text{rms}}X_L=(2)(50)=100\,\text{V}
\]
\( \textbf{Voltage across capacitor:} \)
\[
V_C=I_{\text{rms}}X_C=(2)(20)=40\,\text{V}
\]
\( \textbf{Phasor note:} \) These drops are not simply added arithmetically to get the source voltage because their phases differ.
\( \textbf{Final answer:} \) \(V_R=60\,\text{V}\), \(V_L=100\,\text{V}\), and \(V_C=40\,\text{V}\).
207. The voltages \(V_R\), \(V_L\), and \(V_C\) in a series \(LCR\) circuit cannot generally be added as simple numbers because they
ⓐ. are measured in different units
ⓑ. flow through different branches
ⓒ. are not all in the same phase
ⓓ. are always zero at resonance
Correct Answer: are not all in the same phase
Explanation: In a series \(LCR\) circuit, the same current flows through \(R\), \(L\), and \(C\), but the voltage drops across them have different phase relations with that current. \(V_R\) is in phase with current. \(V_L\) leads current by \(\frac{\pi}{2}\), while \(V_C\) lags current by \(\frac{\pi}{2}\). Since these voltages point in different phasor directions, their resultant must be found by phasor addition. Simple arithmetic addition would ignore the angular separation between the voltage drops.
208. The opposite directions of \(V_L\) and \(V_C\) in a series \(LCR\) phasor diagram occur because
ⓐ. the same current cannot pass through both \(L\) and \(C\)
ⓑ. \(V_L\) leads current and \(V_C\) lags current
ⓒ. the inductor and capacitor always have equal voltages
ⓓ. the resistor voltage cancels both of them
Correct Answer: \(V_L\) leads current and \(V_C\) lags current
Explanation: In a series circuit, current is usually taken as the reference phasor. The inductor voltage \(V_L\) is drawn \(90^\circ\) ahead of current. The capacitor voltage \(V_C\) is drawn \(90^\circ\) behind current. Therefore, \(V_L\) and \(V_C\) lie along the same line but in opposite directions. Their difference, not their arithmetic sum, gives the net reactive voltage.
209. For \(V_R=80\,\text{V}\), \(V_L=120\,\text{V}\), and \(V_C=60\,\text{V}\) in a series \(LCR\) circuit, the rms source voltage is
ⓐ. \(140\,\text{V}\)
ⓑ. \(100\,\text{V}\)
ⓒ. \(200\,\text{V}\)
ⓓ. \(260\,\text{V}\)
Correct Answer: \(100\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(V_R=80\,\text{V}\), \(V_L=120\,\text{V}\), and \(V_C=60\,\text{V}\).
\( \textbf{Required:} \) Resultant rms source voltage \(V\).
\( \textbf{Net reactive voltage:} \)
\[
V_X=V_L-V_C
\]
\( \textbf{Substitution:} \)
\[
V_X=120-60=60\,\text{V}
\]
\( \textbf{Phasor relation:} \)
\[
V=\sqrt{V_R^2+(V_L-V_C)^2}
\]
\( \textbf{Substitution:} \)
\[
V=\sqrt{80^2+60^2}\,\text{V}
\]
\( \textbf{Calculation:} \)
\[
V=\sqrt{6400+3600}=\sqrt{10000}=100\,\text{V}
\]
\( \textbf{Final answer:} \) The rms source voltage is \(100\,\text{V}\).
210. When \(V_L=V_C\) in a series \(LCR\) circuit and \(V_R=50\,\text{V}\), the source voltage at that frequency is
ⓐ. \(0\,\text{V}\)
ⓑ. \(100\,\text{V}\)
ⓒ. \(V_L+V_C+50\,\text{V}\)
ⓓ. \(50\,\text{V}\)
Correct Answer: \(50\,\text{V}\)
Explanation: When \(V_L=V_C\), the inductive and capacitive voltage phasors are equal in magnitude and opposite in direction. Their net reactive voltage is therefore zero. The source voltage phasor then has only the resistive component \(V_R\). Hence the source voltage is equal to \(V_R\), which is \(50\,\text{V}\). This does not mean that \(V_L\) and \(V_C\) are individually absent; they cancel in the phasor sum.
211. Use the arrangement described below.
In a series \(LCR\) circuit, current is taken along the horizontal reference line. The resistor voltage is drawn along this same line. The inductor voltage is drawn upward, and the capacitor voltage is drawn downward.
If \(V_L\gt V_C\), the resultant source voltage lies
ⓐ. above the horizontal reference line
ⓑ. along the horizontal line only
ⓒ. below the horizontal reference line
ⓓ. opposite to the resistor voltage
Correct Answer: above the horizontal reference line
Explanation: With current as reference, \(V_R\) lies horizontally. The inductor voltage \(V_L\) is drawn upward, while the capacitor voltage \(V_C\) is drawn downward. If \(V_L\gt V_C\), the upward reactive component is larger than the downward one. The net reactive voltage is therefore upward. The source voltage is the resultant of horizontal \(V_R\) and upward \(V_L-V_C\), so it lies above the reference line.
212. The impedance of a series \(LCR\) circuit is given by
ⓐ. \(Z=R+X_L+X_C\)
ⓑ. \(Z=\sqrt{R^2+(X_L+X_C)^2}\)
ⓒ. \(Z=\sqrt{R^2+(X_L-X_C)^2}\)
ⓓ. \(Z=R(X_L-X_C)\)
Correct Answer: \(Z=\sqrt{R^2+(X_L-X_C)^2}\)
Explanation: In a series \(LCR\) circuit, resistance \(R\) lies along the current reference direction in the impedance triangle. The net reactance is \(X_L-X_C\), because inductive and capacitive reactances act in opposite phasor directions. The impedance is the resultant of \(R\) and the net reactance. Hence \(Z=\sqrt{R^2+(X_L-X_C)^2}\). Adding \(X_L\) and \(X_C\) directly would ignore their opposite phase effects.
213. Given a series \(LCR\) circuit with \(R=30\,\Omega\), \(X_L=70\,\Omega\), and \(X_C=30\,\Omega\), its impedance is
ⓐ. \(40\,\Omega\)
ⓑ. \(70\,\Omega\)
ⓒ. \(130\,\Omega\)
ⓓ. \(50\,\Omega\)
Correct Answer: \(50\,\Omega\)
Explanation: \( \textbf{Given:} \) \(R=30\,\Omega\), \(X_L=70\,\Omega\), and \(X_C=30\,\Omega\).
\( \textbf{Required:} \) Impedance \(Z\).
\( \textbf{Net reactance:} \)
\[
X=X_L-X_C
\]
\( \textbf{Substitution:} \)
\[
X=70-30=40\,\Omega
\]
\( \textbf{Impedance relation:} \)
\[
Z=\sqrt{R^2+X^2}
\]
\( \textbf{Substitution:} \)
\[
Z=\sqrt{30^2+40^2}\,\Omega
\]
\( \textbf{Calculation:} \)
\[
Z=\sqrt{900+1600}=\sqrt{2500}=50\,\Omega
\]
\( \textbf{Final answer:} \) The impedance is \(50\,\Omega\).
214. For \(R=12\,\Omega\), \(X_L=5\,\Omega\), and \(X_C=21\,\Omega\) in a series \(LCR\) circuit, the impedance is
ⓐ. \(13\,\Omega\)
ⓑ. \(16\,\Omega\)
ⓒ. \(20\,\Omega\)
ⓓ. \(38\,\Omega\)
Correct Answer: \(20\,\Omega\)
Explanation: \( \textbf{Given:} \) \(R=12\,\Omega\), \(X_L=5\,\Omega\), and \(X_C=21\,\Omega\).
\( \textbf{Net reactance:} \)
\[
X=X_L-X_C
\]
\( \textbf{Substitution:} \)
\[
X=5-21=-16\,\Omega
\]
\( \textbf{Magnitude used in impedance:} \) The sign shows capacitive nature, but \(X^2\) enters \(Z\).
\( \textbf{Impedance formula:} \)
\[
Z=\sqrt{R^2+(X_L-X_C)^2}
\]
\( \textbf{Substitution:} \)
\[
Z=\sqrt{12^2+(-16)^2}\,\Omega
\]
\( \textbf{Calculation:} \)
\[
Z=\sqrt{144+256}=\sqrt{400}=20\,\Omega
\]
\( \textbf{Final answer:} \) The impedance is \(20\,\Omega\).
215. The unit of impedance \(Z\) in a series \(LCR\) circuit is
ⓐ. \(\text{A}\)
ⓑ. \(\Omega\)
ⓒ. \(\text{V}\)
ⓓ. \(\text{W}\)
Correct Answer: \(\Omega\)
Explanation: Impedance is the total opposition offered by an \(\text{AC}\) circuit. In a series \(LCR\) circuit, it appears in the relation \(I=\frac{V}{Z}\). Since current is voltage divided by impedance, \(Z\) must have the unit \(\frac{\text{V}}{\text{A}}\). This unit is the ohm, written as \(\Omega\). Even though impedance includes reactance and resistance together, its unit is the same as resistance.
216. If a series \(LCR\) circuit has \(V_{\text{rms}}=200\,\text{V}\) and impedance \(Z=50\,\Omega\), the rms current is
ⓐ. \(4\,\text{A}\)
ⓑ. \(2\,\text{A}\)
ⓒ. \(50\,\text{A}\)
ⓓ. \(10000\,\text{A}\)
Correct Answer: \(4\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(V_{\text{rms}}=200\,\text{V}\), \(Z=50\,\Omega\).
\( \textbf{Required:} \) rms current \(I_{\text{rms}}\).
\( \textbf{Series \(LCR\) current relation:} \)
\[
I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}
\]
\( \textbf{Reason for using \(Z\):} \) The circuit contains resistance and reactance together, so total opposition is impedance.
\( \textbf{Substitution:} \)
\[
I_{\text{rms}}=\frac{200\,\text{V}}{50\,\Omega}
\]
\( \textbf{Calculation:} \)
\[
I_{\text{rms}}=4\,\text{A}
\]
\( \textbf{Unit check:} \) \(\frac{\text{V}}{\Omega}=\text{A}\).
\( \textbf{Final answer:} \) The rms current is \(4\,\text{A}\).
217. With \(R=40\,\Omega\), \(X_L=90\,\Omega\), and \(X_C=60\,\Omega\) in a series \(LCR\) circuit, if \(V_{\text{rms}}=250\,\text{V}\), the rms current is
ⓐ. \(5\,\text{A}\)
ⓑ. \(3\,\text{A}\)
ⓒ. \(4\,\text{A}\)
ⓓ. \(6.25\,\text{A}\)
Correct Answer: \(5\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(R=40\,\Omega\), \(X_L=90\,\Omega\), \(X_C=60\,\Omega\), and \(V_{\text{rms}}=250\,\text{V}\).
\( \textbf{Net reactance:} \)
\[
X=X_L-X_C=90-60=30\,\Omega
\]
\( \textbf{Impedance:} \)
\[
Z=\sqrt{R^2+X^2}
\]
\( \textbf{Substitution:} \)
\[
Z=\sqrt{40^2+30^2}\,\Omega
\]
\( \textbf{Calculation of \(Z\):} \)
\[
Z=\sqrt{1600+900}=\sqrt{2500}=50\,\Omega
\]
\( \textbf{Current relation:} \)
\[
I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}
\]
\( \textbf{Final calculation:} \)
\[
I_{\text{rms}}=\frac{250}{50}=5\,\text{A}
\]
\( \textbf{Final answer:} \) The rms current is \(5\,\text{A}\).
218. The phase angle \(\phi\) in a series \(LCR\) circuit is given by
ⓐ. \(\tan\phi=\frac{R}{X_L-X_C}\)
ⓑ. \(\tan\phi=\frac{X_L-X_C}{R}\)
ⓒ. \(\tan\phi=\frac{X_L+X_C}{R}\)
ⓓ. \(\tan\phi=R(X_L-X_C)\)
Correct Answer: \(\tan\phi=\frac{X_L-X_C}{R}\)
Explanation: In the impedance triangle of a series \(LCR\) circuit, \(R\) is the horizontal side and \(X_L-X_C\) is the vertical reactive side. The phase angle \(\phi\) is the angle between the impedance phasor and the resistance direction. Therefore, \(\tan\phi=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{X_L-X_C}{R}\). The sign of \(X_L-X_C\) tells whether the circuit is inductive or capacitive. Using \(X_L+X_C\) would ignore the opposite phasor directions of \(L\) and \(C\).
219. For \(X_L\gt X_C\) in a series \(LCR\) circuit, the current
ⓐ. leads the source voltage
ⓑ. is always in phase with source voltage
ⓒ. lags the source voltage
ⓓ. becomes zero at every instant
Correct Answer: lags the source voltage
Explanation: If \(X_L\gt X_C\), the net reactance \(X_L-X_C\) is positive. The circuit then behaves inductively. In an inductive circuit, current lags voltage. The resistor part still keeps a component of voltage in phase with current, but the resultant source voltage is ahead of current. The sign of \(X_L-X_C\) determines the lead-lag nature of the series \(LCR\) circuit.
220. For \(X_C\gt X_L\) in a series \(LCR\) circuit, the circuit behaves as
ⓐ. capacitive, with current leading voltage
ⓑ. inductive, with current lagging voltage
ⓒ. purely resistive at all frequencies
ⓓ. an open circuit with zero current always
Correct Answer: capacitive, with current leading voltage
Explanation: When \(X_C\gt X_L\), the net reactance \(X_L-X_C\) is negative. This means the capacitive effect is stronger than the inductive effect. A capacitive circuit has current leading the source voltage. The series circuit still contains \(R\), \(L\), and \(C\), but its net phase behaviour is capacitive. The sign of the net reactance tells the circuit nature, while the magnitude contributes to impedance.
