401. In a generator coil with \(200\) turns rotating in a uniform magnetic field, if the peak flux through each turn is \(2.5\times10^{-3}\,\text{Wb}\) and \(\omega=40\,\text{rad s}^{-1}\), the peak emf is
ⓐ. \(500\,\text{V}\)
ⓑ. \(0.20\,\text{V}\)
ⓒ. \(20\,\text{V}\)
ⓓ. \(8.0\,\text{V}\)
Correct Answer: \(20\,\text{V}\)
Explanation: \( \textbf{Number of turns:} \) \(N=200\).
\( \textbf{Peak flux per turn:} \) \(\phi_0=2.5\times10^{-3}\,\text{Wb}\).
\( \textbf{Angular speed:} \) \(\omega=40\,\text{rad s}^{-1}\).
\( \textbf{Peak flux linkage:} \)
\[
(N\phi)_0=N\phi_0
\]
\[
(N\phi)_0=200(2.5\times10^{-3})=0.50\,\text{Wb}
\]
\( \textbf{Peak emf relation:} \)
\[
\varepsilon_0=(N\phi)_0\omega
\]
\( \textbf{Substitution:} \)
\[
\varepsilon_0=(0.50)(40)
\]
\( \textbf{Final answer:} \) The peak emf is \(20\,\text{V}\).
The calculation must use peak flux linkage, not just the flux through one turn.
402. A generator coil rotates from a position where its area vector is parallel to \(\vec{B}\) to a position where its plane is parallel to \(\vec{B}\). During this quarter turn, the magnitude of flux through the coil
ⓐ. increases from zero to maximum
ⓑ. decreases from maximum to zero
ⓒ. remains maximum throughout
ⓓ. remains zero throughout
Correct Answer: decreases from maximum to zero
Explanation: When the area vector is parallel to \(\vec{B}\), the angle \(\theta\) between them is \(0^\circ\), so the flux has maximum magnitude. When the plane of the coil is parallel to \(\vec{B}\), the area vector is perpendicular to \(\vec{B}\). Then \(\theta=90^\circ\), and the flux becomes zero. In a generator, this changing flux is what produces induced emf. The quarter-turn description helps distinguish the plane of the coil from the area vector direction.
403. Study the comparison of contact arrangements in simple generators.
| Row | Contact arrangement | Usual role |
|---|
| P | Slip rings | Deliver alternating output from a rotating coil |
| Q | Split-ring commutator | Helps make external output unidirectional in a simple \(DC\) generator arrangement |
| R | Brushes | Provide external contact with rotating rings or commutator |
| S | Slip rings | Mechanically reverse coil connections every half turn |
The row with the faulty statement is
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row P
ⓓ. Row S
Correct Answer: Row S
Explanation: Slip rings rotate with the coil and keep each coil end connected to its corresponding external brush. They do not reverse the external connections every half turn, so the output remains alternating. A split-ring commutator is the arrangement that reverses contact every half turn in a simple \(DC\) generator arrangement. Brushes provide sliding electrical contact in both types of machines. Row S incorrectly assigns the commutator action to slip rings.
404. In a simple \(AC\) generator, reversing the direction of rotation while keeping the same magnetic field reverses the generated emf at a given corresponding coil position because
ⓐ. the area of the coil becomes zero permanently
ⓑ. the coil loses all turns
ⓒ. the rate of change of flux changes sign
ⓓ. the magnetic field becomes non-uniform automatically
Correct Answer: the rate of change of flux changes sign
Explanation: The generated emf is given by \(\varepsilon=-\frac{d(N\phi)}{dt}\). Reversing the rotation reverses how the angle between \(\vec{B}\) and \(\vec{A}\) changes with time. For a corresponding position, the sign of the flux-change rate is reversed. Therefore the induced emf direction is also reversed. The reversal comes from changing the motion, not from changing the number of turns or destroying the magnetic field.
405. For a generator coil with \(N=150\), \(A=3.0\times10^{-3}\,\text{m}^2\), \(B=0.40\,\text{T}\), and \(\omega=60\,\text{rad s}^{-1}\), if \(\sin\omega t=\frac{1}{2}\), the instantaneous emf is
ⓐ. \(5.4\,\text{V}\)
ⓑ. \(10.8\,\text{V}\)
ⓒ. \(1.8\,\text{V}\)
ⓓ. \(36\,\text{V}\)
Correct Answer: \(5.4\,\text{V}\)
Explanation: \( \textbf{Given:} \) \(N=150\), \(A=3.0\times10^{-3}\,\text{m}^2\), \(B=0.40\,\text{T}\), \(\omega=60\,\text{rad s}^{-1}\), and \(\sin\omega t=\frac{1}{2}\).
\( \textbf{Generator emf relation:} \)
\[
\varepsilon=NBA\omega\sin\omega t
\]
\( \textbf{Find the peak factor:} \)
\[
NBA\omega=(150)(0.40)(3.0\times10^{-3})(60)
\]
\( \textbf{Intermediate multiplication:} \)
\[
150(3.0\times10^{-3})=0.45
\]
\[
(0.45)(0.40)(60)=10.8
\]
\( \textbf{Instantaneous emf:} \)
\[
\varepsilon=(10.8)\left(\frac{1}{2}\right)
\]
\( \textbf{Final answer:} \) The instantaneous emf is \(5.4\,\text{V}\).
406. Assertion: Increasing the number of turns in a generator coil increases its peak emf if \(B\), \(A\), and \(\omega\) remain unchanged.
Reason: The peak emf is proportional to the flux linkage change rate, and flux linkage contains the factor \(N\).
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, but Reason does not explain Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because the peak emf of a rotating coil is \(\varepsilon_0=NBA\omega\). If \(B\), \(A\), and \(\omega\) are fixed, increasing \(N\) increases \(\varepsilon_0\) directly. The Reason is also true because Faraday's law acts on the rate of change of flux linkage \(N\phi\). More turns give more total flux linkage for the same flux per turn. The Reason explains why the turn factor appears in the generator emf.
407. For a uniformly rotating generator coil, the flux linkage and emf are compared at four instants.
| Instant | \(\omega t\) | Flux linkage \(N\phi=NBA\cos\omega t\) | Emf \(\varepsilon=NBA\omega\sin\omega t\) |
|---|
| P | \(0^\circ\) | Maximum positive | Zero |
| Q | \(90^\circ\) | Zero | Maximum positive |
| R | \(180^\circ\) | Maximum negative | Zero |
| S | \(270^\circ\) | Zero | Maximum positive |
The row with the faulty emf sign is
ⓐ. Row R
ⓑ. Row P
ⓒ. Row Q
ⓓ. Row S
Correct Answer: Row S
Explanation: At \(\omega t=0^\circ\), \(\cos0^\circ=1\) and \(\sin0^\circ=0\), so row P is correct. At \(90^\circ\), flux linkage is zero and \(\sin90^\circ=1\), so row Q is correct. At \(180^\circ\), flux linkage is maximum negative and \(\sin180^\circ=0\), so row R is correct. At \(270^\circ\), \(\sin270^\circ=-1\), so the emf should be maximum negative, not maximum positive. The sign follows from the sine term in the chosen expression.
408. A simple generator is rotated faster, but the magnetic field strength and coil dimensions are unchanged. The maximum magnetic flux through each turn
ⓐ. doubles
ⓑ. unchanged
ⓒ. becomes half
ⓓ. becomes zero
Correct Answer: unchanged
Explanation: The maximum flux through one turn is \(\phi_0=BA\) when the area vector is parallel to the magnetic field. This maximum depends on magnetic field strength and coil area. It does not depend on the angular speed of rotation. Rotating faster increases the rate of change of flux and hence increases peak emf. The generator can produce a larger emf without increasing the maximum flux value itself.
409. A generator produces an emf \(\varepsilon=\varepsilon_0\sin\omega t\). The time interval between two successive positive peaks is
ⓐ. \(\frac{\pi}{2\omega}\)
ⓑ. \(\frac{2\pi}{\omega}\)
ⓒ. \(\frac{\omega}{2\pi}\)
ⓓ. \(\frac{\pi}{\omega}\)
Correct Answer: \(\frac{2\pi}{\omega}\)
Explanation: Successive positive peaks of \(\sin\omega t\) occur after one full cycle. The angular phase must increase by \(2\pi\) between one positive peak and the next. Since phase is \(\omega t\), the time period is given by \(\omega T=2\pi\). Thus \(T=\frac{2\pi}{\omega}\). The interval \(\frac{\pi}{\omega}\) would take the emf from a positive peak to a negative peak, not to the next positive peak.
410. The emf of a generator coil is described by \(\varepsilon=12\sin(50t)\,\text{V}\), with peak emf \(12\,\text{V}\). At \(t=0\), the flux linkage is most consistent with
ⓐ. impossible to relate to flux linkage
ⓑ. zero and increasing positively
ⓒ. maximum positive flux linkage
ⓓ. maximum negative only
Correct Answer: maximum positive flux linkage
Explanation: If \(\varepsilon=12\sin(50t)\), then the emf is zero at \(t=0\) and becomes positive just after \(t=0\). With Faraday's law, this is consistent with a flux linkage that starts at maximum positive and then decreases. The form \(N\phi=(N\phi)_0\cos(50t)\) has maximum positive value at \(t=0\). Its derivative is negative just after \(t=0\), so \(\varepsilon=-\frac{d(N\phi)}{dt}\) becomes positive. The flux and emf are phase-shifted because emf depends on slope of flux.
411. A \(DC\) generator arrangement uses a split-ring commutator. The commutator makes the external current unidirectional by
ⓐ. keeping the magnetic flux constant in the coil
ⓑ. stopping the coil after every half turn
ⓒ. removing the induced emf inside the coil
ⓓ. reverses external connections every half turn
Correct Answer: reverses external connections every half turn
Explanation: The emf induced in a rotating coil reverses direction every half turn. In a simple \(DC\) generator arrangement, the split-ring commutator reverses the external connections at the same time. This makes the current in the external circuit remain in one direction, although it may pulsate in magnitude. The commutator does not remove induction or make the flux constant. It changes the way the internally alternating emf is delivered to the external circuit.
412. A simple \(AC\) generator and a simple \(DC\) generator can have the same rotating coil and magnetic field. The main constructional difference in the output connection is
ⓐ. AC uses slip rings; DC uses a split-ring commutator
ⓑ. a chemical cell in the \(AC\) generator and a magnet in the \(DC\) generator
ⓒ. zero magnetic flux in the \(AC\) generator and changing flux in the \(DC\) generator
ⓓ. no brushes in the \(AC\) generator and no coil in the \(DC\) generator
Correct Answer: AC uses slip rings; DC uses a split-ring commutator
Explanation: Both simple generator types can use electromagnetic induction in a rotating coil. The induced emf inside the coil is alternating because the flux-change direction reverses every half turn. In an \(AC\) generator, slip rings deliver this alternating emf to the external circuit. In a simple \(DC\) generator arrangement, a split-ring commutator reverses the connections every half turn to make the external current unidirectional. The difference is therefore in the external connection mechanism, not in abandoning flux change.
413. In a simple \(DC\) generator arrangement, the emf induced in the rotating coil is internally alternating, but the external output becomes unidirectional because
ⓐ. the brushes remove the induced emf from the circuit
ⓑ. the magnetic field stops changing after every half turn
ⓒ. the coil resistance becomes zero whenever the emf reverses
ⓓ. commutator reverses connections every half turn
Correct Answer: commutator reverses connections every half turn
Explanation: A rotating coil in a magnetic field has changing flux, so the induced emf in the coil reverses after every half turn. A split-ring commutator reverses the connection of the coil ends to the external circuit at the same instant. This makes the current through the external load remain in the same direction. The output is not perfectly steady; it is usually pulsating unidirectional output in the simple model. The commutator changes the external connection, not the basic Faraday-law production of emf inside the coil.
414. A simple generator coil rotates at \(25\) revolutions per second. For a two-pole generator, the frequency of the generated alternating emf is
ⓐ. \(50\,\text{Hz}\)
ⓑ. \(25\,\text{Hz}\)
ⓒ. \(12.5\,\text{Hz}\)
ⓓ. \(100\,\text{Hz}\)
Correct Answer: \(25\,\text{Hz}\)
Explanation: \( \textbf{Rotational speed:} \) The coil completes \(25\) revolutions each second.
\( \textbf{Flux cycle idea:} \) In a simple two-pole generator, one complete revolution gives one complete cycle of flux variation.
\( \textbf{Emf cycle:} \) Since emf is linked to the rate of change of flux, it also completes one cycle per revolution.
\( \textbf{Frequency relation:} \)
\[
f=25\,\text{s}^{-1}
\]
\( \textbf{Unit conversion:} \)
\[
1\,\text{s}^{-1}=1\,\text{Hz}
\]
\( \textbf{Final answer:} \) The generated emf frequency is \(25\,\text{Hz}\).
For this simple case, the electrical frequency equals the mechanical revolutions per second.
415. For a generator coil rotating with angular speed \(\omega\), the relation between angular speed and frequency of the generated emf is
ⓐ. \(f=\frac{\omega}{2\pi}\)
ⓑ. \(f=\frac{2\pi}{\omega}\)
ⓒ. \(f=\omega^2\)
ⓓ. \(f=2\pi\omega\)
Correct Answer: \(f=\frac{\omega}{2\pi}\)
Explanation: Angular speed \(\omega\) measures how many radians are covered each second. One complete cycle corresponds to an angular change of \(2\pi\,\text{rad}\). Therefore the number of complete cycles per second is \(\frac{\omega}{2\pi}\). This is the frequency \(f\). The expression \(\frac{2\pi}{\omega}\) gives the time period, not the frequency.
416. A generator produces \(\varepsilon=20\sin(100\pi t)\,\text{V}\). Its frequency is
ⓐ. \(200\pi\,\text{Hz}\)
ⓑ. \(100\,\text{Hz}\)
ⓒ. \(50\,\text{Hz}\)
ⓓ. \(20\,\text{Hz}\)
Correct Answer: \(50\,\text{Hz}\)
Explanation: \( \textbf{Given emf:} \)
\[
\varepsilon=20\sin(100\pi t)\,\text{V}
\]
\( \textbf{Compare with standard form:} \)
\[
\varepsilon=\varepsilon_0\sin\omega t
\]
\( \textbf{Angular frequency:} \)
\[
\omega=100\pi\,\text{rad s}^{-1}
\]
\( \textbf{Frequency relation:} \)
\[
f=\frac{\omega}{2\pi}
\]
\( \textbf{Substitution:} \)
\[
f=\frac{100\pi}{2\pi}
\]
\( \textbf{Final answer:} \) The frequency is \(50\,\text{Hz}\).
The coefficient of \(t\) in the sine function is angular frequency, so it must be divided by \(2\pi\) to get frequency.
417. A rotating generator coil has peak emf \(\varepsilon_0\). Its rms emf for sinusoidal output is
ⓐ. \(\varepsilon_0/\sqrt{2}\)
ⓑ. \(\sqrt{2}\varepsilon_0\)
ⓒ. \(\varepsilon_0/\sqrt{3}\)
ⓓ. \(\sqrt{3}\varepsilon_0\)
Correct Answer: \(\varepsilon_0/\sqrt{2}\)
Explanation: For a sinusoidal alternating emf, the rms value is the steady \(DC\) value that would produce the same heating effect in a resistor. If the peak value is \(\varepsilon_0\), the rms value is \(\varepsilon_{\text{rms}}=\frac{\varepsilon_0}{\sqrt{2}}\). The rms value is smaller than the peak value because the sine wave is not at its maximum throughout the cycle. The factor \(\sqrt{2}\) is specific to a sinusoidal waveform. This relation should not be used for a non-sinusoidal waveform unless its rms value is known separately.
418. A generator has peak emf \(120\,\text{V}\) and supplies a sinusoidal output. The rms emf is closest to
ⓐ. \(85\,\text{V}\)
ⓑ. \(60\,\text{V}\)
ⓒ. \(240\,\text{V}\)
ⓓ. \(170\,\text{V}\)
Correct Answer: \(85\,\text{V}\)
Explanation: \( \textbf{Peak emf:} \)
\[
\varepsilon_0=120\,\text{V}
\]
\( \textbf{Rms relation for sinusoidal emf:} \)
\[
\varepsilon_{\text{rms}}=\frac{\varepsilon_0}{\sqrt{2}}
\]
\( \textbf{Substitution:} \)
\[
\varepsilon_{\text{rms}}=\frac{120}{1.414}
\]
\( \textbf{Calculation:} \)
\[
\varepsilon_{\text{rms}}\approx84.9\,\text{V}
\]
\( \textbf{Final answer:} \) The rms emf is closest to \(85\,\text{V}\).
The rms value is the heating-equivalent value, not the maximum instantaneous value.
419. A generator coil is changed from \(100\) turns to \(200\) turns, and its angular speed is reduced to half. If \(B\) and \(A\) remain unchanged, the peak emf
ⓐ. becomes four times larger
ⓑ. becomes double
ⓒ. becomes half
ⓓ. remains unchanged
Correct Answer: remains unchanged
Explanation: The peak emf of a rotating generator coil is \(\varepsilon_0=NBA\omega\). Doubling \(N\) multiplies \(\varepsilon_0\) by \(2\). Reducing \(\omega\) to half multiplies \(\varepsilon_0\) by \(\frac{1}{2}\). The combined factor is \(2\times\frac{1}{2}=1\). Therefore the peak emf remains unchanged, although the frequency becomes smaller because the angular speed is lower.
420. For a generator coil rotating in a magnetic field, the flux linkage is \(N\phi=(0.050)\cos(200t)\,\text{Wb}\). The generated emf at \(t=\frac{\pi}{800}\,\text{s}\) is closest to
ⓐ. \(0\,\text{V}\)
ⓑ. \(5.0\,\text{V}\)
ⓒ. \(7.1\,\text{V}\)
ⓓ. \(-10\,\text{V}\)
Correct Answer: \(7.1\,\text{V}\)
Explanation: \( \textbf{Flux linkage:} \)
\[
N\phi=0.050\cos(200t)\,\text{Wb}
\]
\( \textbf{Faraday's law:} \)
\[
\varepsilon=-\frac{d(N\phi)}{dt}
\]
\( \textbf{Differentiate:} \)
\[
\frac{d(N\phi)}{dt}=-0.050(200)\sin(200t)
\]
\[
\varepsilon=10\sin(200t)
\]
\( \textbf{Given time:} \)
\[
t=\frac{\pi}{800}\,\text{s}
\]
\( \textbf{Phase angle:} \)
\[
200t=200\left(\frac{\pi}{800}\right)=\frac{\pi}{4}
\]
\( \textbf{Substitution:} \)
\[
\varepsilon=10\sin\frac{\pi}{4}=10\left(\frac{1}{\sqrt{2}}\right)
\]
\[
\varepsilon\approx7.1\,\text{V}
\]
\( \textbf{Final answer:} \) The generated emf is closest to \(7.1\,\text{V}\).
