101. A voltage waveform reaches its positive maximum earlier than the current waveform in the same circuit. The voltage is then said to
ⓐ. lag the current
ⓑ. be equal to zero throughout
ⓒ. lead the current
ⓓ. have no frequency
Correct Answer: lead the current
Explanation: Lead-lag language compares the timing of corresponding points on two waveforms. If the voltage reaches its positive maximum earlier, it is ahead in phase. The quantity that reaches the same phase condition earlier is said to lead. The current reaches its maximum later, so current lags voltage in this situation. Lead and lag describe phase timing, not whether the quantity exists or has frequency.
102. For two sinusoidal quantities of the same frequency, a time difference \(\Delta t\) corresponds to a phase difference
ⓐ. \(\Delta\phi=\frac{\Delta t}{\omega}\)
ⓑ. \(\Delta\phi=\omega+\Delta t\)
ⓒ. \(\Delta\phi=\frac{1}{\omega\Delta t}\)
ⓓ. \(\Delta\phi=\omega\Delta t\)
Correct Answer: \(\Delta\phi=\omega\Delta t\)
Explanation: Phase angle changes at the rate \(\omega\), where \(\omega\) is angular frequency. If two corresponding points on two waveforms are separated by a time interval \(\Delta t\), the corresponding phase difference is \(\Delta\phi=\omega\Delta t\). The units also support this relation because \((\text{rad s}^{-1})(\text{s})=\text{rad}\). Adding \(\omega\) and \(\Delta t\) is not meaningful because their units are different. The relation converts a time shift into an angular shift.
103. An \(\text{AC}\) voltage and current have the same frequency \(50\,\text{Hz}\). The current reaches its positive maximum \(5.0\,\text{ms}\) after the voltage. The phase difference is
ⓐ. \(\frac{\pi}{4}\,\text{rad}\)
ⓑ. \(\pi\,\text{rad}\)
ⓒ. \(\frac{\pi}{2}\,\text{rad}\)
ⓓ. \(2\pi\,\text{rad}\)
Correct Answer: \(\frac{\pi}{2}\,\text{rad}\)
Explanation: \( \textbf{Given:} \) \(f=50\,\text{Hz}\), \(\Delta t=5.0\,\text{ms}=5.0\times10^{-3}\,\text{s}\).
\( \textbf{Required:} \) Phase difference \(\Delta\phi\).
\( \textbf{Angular frequency:} \)
\[
\omega=2\pi f
\]
\( \textbf{Substitution for \(\omega\):} \)
\[
\omega=2\pi(50)=100\pi\,\text{rad s}^{-1}
\]
\( \textbf{Time-to-phase relation:} \)
\[
\Delta\phi=\omega\Delta t
\]
\( \textbf{Substitution:} \)
\[
\Delta\phi=(100\pi)(5.0\times10^{-3})
\]
\( \textbf{Calculation:} \)
\[
\Delta\phi=0.5\pi=\frac{\pi}{2}\,\text{rad}
\]
\( \textbf{Direction meaning:} \) Since current reaches the maximum later, current lags voltage by \(\frac{\pi}{2}\).
\( \textbf{Final answer:} \) The phase difference is \(\frac{\pi}{2}\,\text{rad}\).
104. Use the graph description below.
Two sinusoidal graphs have the same time period. Graph P crosses zero with positive slope at \(t=0\). Graph Q crosses zero with positive slope at a later time.
The suitable conclusion is that
ⓐ. Graph Q leads graph P
ⓑ. Graph Q lags graph P
ⓒ. the two graphs must have different frequencies
ⓓ. the two graphs must have zero amplitude
Correct Answer: Graph Q lags graph P
Explanation: Both graphs have the same time period, so they have the same frequency. The comparison is therefore about phase, not frequency. Graph P reaches the same reference condition earlier, because it crosses zero with positive slope at \(t=0\). Graph Q reaches that condition later, so Q lags P. A delayed occurrence of the same waveform feature is the graphical sign of lag.
105. A sinusoidal current lags a voltage by \(\frac{\pi}{3}\). If the voltage is written as \(v=V_0\sin\omega t\), a suitable current expression is
ⓐ. \(i=I_0\sin(\omega t-\frac{\pi}{3})\)
ⓑ. \(i=I_0\sin(\omega t+\frac{\pi}{3})\)
ⓒ. \(i=I_0\sin\frac{\omega t}{3}\)
ⓓ. \(i=I_0\sin(\frac{\pi}{3}-\omega t)\)
Correct Answer: \(i=I_0\sin(\omega t-\frac{\pi}{3})\)
Explanation: A lagging current reaches the same phase condition later than the voltage. If the voltage is taken as \(v=V_0\sin\omega t\), a current lag of \(\frac{\pi}{3}\) is represented by subtracting that phase from the argument of the current. Thus the current may be written as \(i=I_0\sin(\omega t-\frac{\pi}{3})\). A plus sign would represent current leading the voltage by \(\frac{\pi}{3}\). The phase shift is placed inside the sine argument because it changes timing, not amplitude.
106. Consider the following statements about phase difference in \(\text{AC}\).
I. Phase difference may be expressed in radians.
II. A phase difference of \(0\) means the two quantities are in phase.
III. If current leads voltage, current reaches corresponding maxima earlier than voltage.
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II, and III
Correct Answer: I, II, and III
Explanation: Statement I is true because phase is an angular quantity and is commonly expressed in radians. Statement II is true because zero phase difference means corresponding points of the two sinusoidal quantities occur together. Statement III is also true because the leading quantity reaches the same phase condition earlier. Lead-lag language is a timing description for quantities of the same frequency. The statements together describe the angle, zero-phase case, and graphical meaning of lead.
107. Two \(50\,\text{Hz}\) sinusoidal quantities have a phase difference of \(\frac{\pi}{6}\). The corresponding time difference between them is
ⓐ. \(\frac{1}{300}\,\text{s}\)
ⓑ. \(\frac{1}{600}\,\text{s}\)
ⓒ. \(\frac{1}{100}\,\text{s}\)
ⓓ. \(\frac{1}{50}\,\text{s}\)
Correct Answer: \(\frac{1}{600}\,\text{s}\)
Explanation: \( \textbf{Given:} \) \(f=50\,\text{Hz}\), \(\Delta\phi=\frac{\pi}{6}\).
\( \textbf{Required:} \) Time difference \(\Delta t\).
\( \textbf{Angular frequency:} \)
\[
\omega=2\pi f=2\pi(50)=100\pi\,\text{rad s}^{-1}
\]
\( \textbf{Relation between phase and time difference:} \)
\[
\Delta\phi=\omega\Delta t
\]
\( \textbf{Rearranging:} \)
\[
\Delta t=\frac{\Delta\phi}{\omega}
\]
\( \textbf{Substitution:} \)
\[
\Delta t=\frac{\pi/6}{100\pi}
\]
\( \textbf{Simplification:} \)
\[
\Delta t=\frac{1}{600}\,\text{s}
\]
\( \textbf{Final answer:} \) The time difference is \(\frac{1}{600}\,\text{s}\).
108. A phasor for voltage is ahead of the phasor for current by an angle \(\phi\) in a same-frequency \(\text{AC}\) circuit. This means
ⓐ. current leads voltage by \(\phi\)
ⓑ. voltage and current are necessarily in phase
ⓒ. the current must be zero at all instants
ⓓ. voltage leads current by \(\phi\)
Correct Answer: voltage leads current by \(\phi\)
Explanation: In a phasor diagram, the phasor that is ahead in the direction of rotation represents the quantity that leads in phase. If the voltage phasor is ahead of the current phasor by \(\phi\), voltage leads current by \(\phi\). The current then lags voltage by the same angle. This does not mean the two quantities are in phase, because in-phase quantities have zero angular separation. A phase difference changes relative timing, not the existence of current.
109. In a pure resistive \(\text{AC}\) circuit, the instantaneous voltage and current are related by
ⓐ. \(v=iR\)
ⓑ. \(v=L\frac{di}{dt}\)
ⓒ. \(i=C\frac{dv}{dt}\)
ⓓ. \(v=\frac{i}{\omega C}\)
Correct Answer: \(v=iR\)
Explanation: A pure resistive \(\text{AC}\) circuit contains only resistance \(R\), so the voltage-current relation follows Ohm’s law at every instant. The instantaneous voltage is proportional to instantaneous current, giving \(v=iR\). This does not mean that \(v\) and \(i\) are constant; both may vary sinusoidally with time. It means their ratio remains \(R\) at each instant. Relations involving \(L\) or \(C\) belong to inductive or capacitive circuits, not to a pure resistor.
110. A resistor \(R\) is connected to a sinusoidal voltage \(v=V_0\sin\omega t\). The current through the resistor is
ⓐ. \(i=\frac{R}{V_0}\sin\omega t\)
ⓑ. \(i=\frac{V_0}{R}\sin\omega t\)
ⓒ. \(i=V_0R\sin\omega t\)
ⓓ. \(i=\frac{V_0}{\omega R}\cos\omega t\)
Correct Answer: \(i=\frac{V_0}{R}\sin\omega t\)
Explanation: In a pure resistor, the instantaneous relation is \(v=iR\). If \(v=V_0\sin\omega t\), then \(i=\frac{v}{R}\). Substituting the given voltage gives \(i=\frac{V_0}{R}\sin\omega t\). The current has the same sine factor as the voltage, so there is no phase shift between them. The peak current is therefore \(I_0=\frac{V_0}{R}\).
111. In a pure resistor connected to a sinusoidal \(\text{AC}\) source, voltage and current are
ⓐ. in phase
ⓑ. out of phase by \(\frac{\pi}{2}\), with current lagging
ⓒ. out of phase by \(\frac{\pi}{2}\), with current leading
ⓓ. always opposite in phase by \(\pi\)
Correct Answer: in phase
Explanation: For a pure resistor, \(v=iR\) at every instant. Since \(R\) is a positive constant, voltage and current rise, fall, pass through zero, and reach their positive and negative peaks at the same times. This means the phase difference between them is \(0\). A \(\frac{\pi}{2}\) phase difference is associated with ideal inductive or capacitive behaviour, not pure resistance. The resistor changes the size of current according to \(R\), but it does not shift its phase.
112. A pure resistor of resistance \(20\,\Omega\) is connected to an \(\text{AC}\) source with \(V_0=100\,\text{V}\). The peak current is
ⓐ. \(2\,\text{A}\)
ⓑ. \(20\,\text{A}\)
ⓒ. \(5\,\text{A}\)
ⓓ. \(2000\,\text{A}\)
Correct Answer: \(5\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(R=20\,\Omega\), \(V_0=100\,\text{V}\).
\( \textbf{Required:} \) Peak current \(I_0\).
\( \textbf{Relation for pure resistor:} \)
\[
I_0=\frac{V_0}{R}
\]
\( \textbf{Why this relation applies:} \) In a pure resistor, Ohm’s law applies to peak values as well as instantaneous values because voltage and current are in phase.
\( \textbf{Substitution:} \)
\[
I_0=\frac{100\,\text{V}}{20\,\Omega}
\]
\( \textbf{Calculation:} \)
\[
I_0=5\,\text{A}
\]
\( \textbf{Unit check:} \) \(\frac{\text{V}}{\Omega}=\text{A}\).
\( \textbf{Final answer:} \) The peak current is \(5\,\text{A}\).
113. A resistor is supplied with \(V_{\text{rms}}=220\,\text{V}\) and carries \(I_{\text{rms}}=2\,\text{A}\). Its resistance is
ⓐ. \(55\,\Omega\)
ⓑ. \(110\,\Omega\)
ⓒ. \(220\,\Omega\)
ⓓ. \(440\,\Omega\)
Correct Answer: \(110\,\Omega\)
Explanation: \( \textbf{Given:} \) \(V_{\text{rms}}=220\,\text{V}\), \(I_{\text{rms}}=2\,\text{A}\).
\( \textbf{Required:} \) Resistance \(R\).
\( \textbf{Pure resistor relation:} \)
\[
I_{\text{rms}}=\frac{V_{\text{rms}}}{R}
\]
\( \textbf{Rearranging:} \)
\[
R=\frac{V_{\text{rms}}}{I_{\text{rms}}}
\]
\( \textbf{Substitution:} \)
\[
R=\frac{220\,\text{V}}{2\,\text{A}}
\]
\( \textbf{Calculation:} \)
\[
R=110\,\Omega
\]
\( \textbf{Interpretation:} \) The rms values can be used directly in Ohm’s law for a pure resistor.
\( \textbf{Final answer:} \) The resistance is \(110\,\Omega\).
114. Use the graph description below.
A voltage-time graph and a current-time graph for a circuit cross zero together, reach positive maxima together, and reach negative maxima together.
The circuit behaviour most directly shown by these two graphs is
ⓐ. pure resistive behaviour
ⓑ. pure inductive behaviour
ⓒ. pure capacitive behaviour
ⓓ. transformer action
Correct Answer: pure resistive behaviour
Explanation: If voltage and current cross zero and reach maxima at the same instants, their phase difference is \(0\). A pure resistor has voltage and current in phase because \(v=iR\). In a pure inductor, current lags voltage by \(\frac{\pi}{2}\). In a pure capacitor, current leads voltage by \(\frac{\pi}{2}\). The described timing matches resistive behaviour rather than energy-storage-only behaviour.
115. A phasor diagram for a pure resistor is drawn using current as the reference phasor. The voltage phasor should be drawn
ⓐ. \(90^\circ\) ahead of the current phasor
ⓑ. \(90^\circ\) behind the current phasor
ⓒ. opposite to the current phasor
ⓓ. along the current phasor
Correct Answer: along the current phasor
Explanation: In a pure resistor, voltage and current are in phase. When current is chosen as the reference phasor, an in-phase voltage phasor must lie along the same line. Since the resistance \(R\) is positive, \(v\) and \(i\) have the same sign at the same instant. A \(90^\circ\) shift would indicate inductive or capacitive phase behaviour. The phasor diagram for a pure resistor has no angular separation between \(V\) and \(I\).
116. A pure resistor has \(R=40\,\Omega\) and is connected to \(v=120\sin\omega t\,\text{V}\). The current equation is
ⓐ. \(i=3\sin\omega t\,\text{A}\)
ⓑ. \(i=40\sin\omega t\,\text{A}\)
ⓒ. \(i=120\sin(\omega t-\frac{\pi}{2})\,\text{A}\)
ⓓ. \(i=4800\sin\omega t\,\text{A}\)
Correct Answer: \(i=3\sin\omega t\,\text{A}\)
Explanation: \( \textbf{Given voltage:} \) \(v=120\sin\omega t\,\text{V}\).
\( \textbf{Given resistance:} \) \(R=40\,\Omega\).
\( \textbf{Instantaneous relation in a resistor:} \)
\[
v=iR
\]
\( \textbf{Current expression:} \)
\[
i=\frac{v}{R}
\]
\( \textbf{Substitution:} \)
\[
i=\frac{120\sin\omega t}{40}\,\text{A}
\]
\( \textbf{Simplification:} \)
\[
i=3\sin\omega t\,\text{A}
\]
\( \textbf{Phase check:} \) The sine argument remains \(\omega t\), so current is in phase with voltage.
\( \textbf{Final answer:} \) The current is \(i=3\sin\omega t\,\text{A}\).
117. In an \(\text{AC}\) circuit, the instantaneous voltage and current at a certain instant are \(v=12\,\text{V}\) and \(i=-3\,\text{A}\). The instantaneous power at that instant is
ⓐ. \(-36\,\text{W}\)
ⓑ. \(+36\,\text{W}\)
ⓒ. \(-4\,\Omega\)
ⓓ. \(15\,\text{W}\)
Correct Answer: \(-36\,\text{W}\)
Explanation: \( \textbf{Given:} \) \(v=12\,\text{V}\), \(i=-3\,\text{A}\).
\( \textbf{Required:} \) Instantaneous power \(p\).
\( \textbf{Formula:} \)
\[
p=vi
\]
\( \textbf{Substitution:} \)
\[
p=(12)(-3)
\]
\( \textbf{Calculation:} \)
\[
p=-36\,\text{W}
\]
\( \textbf{Sign meaning:} \) The negative sign shows that, at this instant, energy is being returned by the circuit element instead of being absorbed by it.
\( \textbf{Final answer:} \) The instantaneous power is \(-36\,\text{W}\).
118. In a pure resistor carrying sinusoidal \(\text{AC}\), the instantaneous power is never negative because
ⓐ. voltage and current always have opposite signs
ⓑ. current is zero throughout the cycle
ⓒ. voltage and current are in phase
ⓓ. resistance changes sign every half cycle
Correct Answer: voltage and current are in phase
Explanation: In a pure resistor, voltage and current are in phase. During the positive half cycle, both \(v\) and \(i\) are positive, so \(p=vi\) is positive. During the negative half cycle, both \(v\) and \(i\) are negative, so their product is again positive. At zero crossings, instantaneous power becomes zero. A resistor absorbs energy from the source in both half cycles rather than returning energy as an ideal inductor or capacitor does.
119. If \(v=V_0\sin\omega t\) and \(i=I_0\sin\omega t\) in a pure resistor, the instantaneous power is
ⓐ. \(p=V_0I_0\sin\omega t\)
ⓑ. \(p=\frac{V_0}{I_0}\sin^2\omega t\)
ⓒ. \(p=V_0I_0\cos\omega t\)
ⓓ. \(p=V_0I_0\sin^2\omega t\)
Correct Answer: \(p=V_0I_0\sin^2\omega t\)
Explanation: In a pure resistor, voltage and current have the same phase, so they may be written as \(v=V_0\sin\omega t\) and \(i=I_0\sin\omega t\). Instantaneous power is \(p=vi\). Multiplying the two expressions gives \(p=V_0I_0\sin^2\omega t\). Since \(\sin^2\omega t\) is never negative, the instantaneous power is never negative. This matches the physical idea that a pure resistor continuously absorbs energy.
120. A pure resistor has \(V_{\text{rms}}=120\,\text{V}\) and \(I_{\text{rms}}=2.5\,\text{A}\). The average power consumed is
ⓐ. \(48\,\text{W}\)
ⓑ. \(300\,\text{W}\)
ⓒ. \(120\,\text{W}\)
ⓓ. \(600\,\text{W}\)
Correct Answer: \(300\,\text{W}\)
Explanation: \( \textbf{Given:} \) \(V_{\text{rms}}=120\,\text{V}\), \(I_{\text{rms}}=2.5\,\text{A}\).
\( \textbf{Required:} \) Average power \(P\) in a pure resistor.
\( \textbf{Power relation:} \)
\[
P=V_{\text{rms}}I_{\text{rms}}
\]
\( \textbf{Why no phase factor is needed:} \) In a pure resistor, voltage and current are in phase, so \(\cos\phi=1\).
\( \textbf{Substitution:} \)
\[
P=(120)(2.5)\,\text{W}
\]
\( \textbf{Calculation:} \)
\[
P=300\,\text{W}
\]
\( \textbf{Unit check:} \) \(\text{V}\times\text{A}=\text{W}\) for real power in this resistive case.
\( \textbf{Final answer:} \) The average power consumed is \(300\,\text{W}\).
