AC Circuits Question Bank |
\[1\star\]
A series RLC circuit is connected to an AC source. If the capacitive reactance is less than the inductive reactance \[XL > XC \], then one of the following statements is correct:
(θ ) Current leads voltage by -C
(θ) Voltage leads current by -A
(𝜋/2 ) Voltage leads current by -D
Current and voltage are in phase -B
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\[2\star\]
In an LC circuit, a capacitor is charged by a battery and then connected to a pure inductor. When the current in the circuit is at its maximum value, then:
Electrical energy is at maximum -C
Capacitor is fully charged -A
Magnetic energy is at maximum -D
Inductor stores energy and capacitor -B
stores energy
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\[3\star\]
In an oscillating circuit, a capacitor with capacitance \[6 n𝐹\] and charge \[3 𝜇c\] is connected to an inductor with self-inductance coefficient \[0.2 H \]. One of the following values cannot be the current intensity in the circuit:
\[ i = - 0.042\;\; A \;\;\;\;\;\;-C\]
\[ i = 0.092\;\; A\;\;\;\;\;\;-A\]
\[ i = - 0.02 \;\;A \;\;\;\;\;-D\]
\[ i = 0.082 \;\;A\;\;\;\;\;\;-B\]
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\[4\star\]
In an LC circuit, during one cycle, the electrical energy equals the magnetic energy
\[\frac{1}{2}\frac{q^2}{c}=\frac{1}{2}L.i^2\]
Three times -C
Once -A
Four times -D
Twice -B
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\[5\star\]
In an AC circuit containing a single element
(parallel plate capacitor - cylindrical capacitor - inductor - ohmic resistor)
the source frequency was increased and it was observed that the maximum value
of the alternating current did not change. Therefore, the circuit contains:
Inductor -C
Parallel plate capacitor -A
Ohmic resistor -D
Cylindrical capacitor -B
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\[6\star\]
At what frequency does the reactance of an inductor with self-inductance coefficient \[0.4 H \] reach an impedance of \[93 Ω\]?
\[ f = 37\;\; HZ\;\;\;\;\;\;-C\]
\[ f = 35\;\; HZ \;\;\;\;\;\;-A\]
\[f = 38 \;\;HZ \;\;\;\;\;-D\]
\[ f = 36 \;\;HZ\;\;\;\;\;\;-B\]
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\[7\star\]
A capacitive capacitor with capacitance
C= 4.0 × 10–4F
is connected to an AC source
with source voltage equation of the form
\[ V (t) = 110 sin (126t )\]
The maximum current passing through the capacitor equals:
\[ i_{max}=1.42 \;\;A\;\;\;\;\;\;-C\]
\[ i_{max}=5.54 \;\;A \;\;\;\;\;\;-A\]
\[ i_{max}=3.85 \;\;A\;\;\;\;\;-D\]
\[ i_{max}=2.63 \;\;A\;\;\;\;\;\;-B\]
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\[8\star\]
In an AC circuit, the ammeter and voltmeter were monitored
and it was observed that the current lags behind the voltage in phase by
\[𝜑=\frac{𝜋}{2}=90^0\]. Therefore, the circuit contains:
Capacitive capacitor -C
Ohmic resistor -A
None of the above devices -D
Pure inductor -B
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\[9\star\]
In the adjacent figure, an
\[RLC\]
circuit is connected to an AC source with equation \[V(t) = 210 Sin (300 t ) \]. If the self-inductance coefficient of the inductor,
the capacitance of the capacitor, and the ohmic resistance are
\[ L= 0.5 H \;\;\;\;\;\;\;\;C= 5 𝜇F\;\;\;\;\;\;\;\;R=150 Ω\]
then the impedance of the circuit equals:
\[ Z= 335\;\; Ω \;\;\;\;\;\;-C\]
\[ Z= 620\;\; Ω \;\;\;\;\;\;-A\]
\[ Z= 427\;\; Ω\;\;\;\;\;-D\]
\[ Z= 538\;\; Ω \;\;\;\;\;\;-B\]
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\[10\star\]
In the adjacent figure, an
\[RLC\]
circuit is shown. If the maximum voltage
of the source is \[V_m = 200 V \], the maximum voltage across the resistor is \[V_mR = 150 V \], and the maximum voltage
across the inductor is \[V_mL = 180 V \], then the maximum voltage across the capacitor equals:
\[ VC =120 \;\;V\;\;\;\;\;\;-C\]
\[ VC =47.7 \;\;V \;\;\;\;\;\;-A\]
\[ VC =253.2\;\; V \;\;\;\;\;-D\]
\[ VC =93.2\;\; V \;\;\;\;\;\;-B\]
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\[11\star\]
An AC circuit connected to a time-varying EMF source with the equation \[V(t) = 210 Sin (60 𝜋 t ) \] is connected to an inductor with self-inductance \[L=0.2 H \]. The equation of the alternating current representing this circuit is:
\[ 𝐼(t) = 5.57 Sin (60𝜋 t - 𝜋/2 )\;\;\;\;\;\;-C\]
\[ 𝐼(t) = 5.67 Sin (60𝜋 t + 𝜋/2 ) \;\;\;\;\;\;-A\]
\[ 𝐼(t) = 17.5 Sin (60𝜋 t - 𝜋/2 )\;\;\;\;\;-D\]
\[ 𝐼(t) = 17.5 Sin (60𝜋 t + 𝜋/2 ) \;\;\;\;\;\;-B\]
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\[12\star\]
An AC circuit is connected to a time-varying EMF source.
The source voltage equation is \[V(t) = 210 Sin (60 𝜋 t )\].
The source is connected to an ohmic resistor \[R=20 Ω\].
The AC current equation representing this circuit is:
\[ 𝐼(t) = 10.5 Sin (60𝜋 t )\;\;\;\;\;\;-C\]
\[ 𝐼(t) = 10.5 Sin (60𝜋 t+ 𝜋/2 ) \;\;\;\;\;\;-A\]
\[ 𝐼(t) = 14.7 Sin (60𝜋 t +𝜋/2)\;\;\;\;\;-D\]
\[ 𝐼(t) = 14.7 Sin (60𝜋 t -𝜋/2) \;\;\;\;\;\;-B\]
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\[13\star\]
In the adjacent figure, an LC circuit is connected to an AC source with equation \[V(t) = 110 Sin (300 t )\].
If the self-inductance coefficient of the coil is \[L= 0.4 H\] and the capacitor capacity is \[ C= 9 𝜇F \],
then the impedance value of the circuit and the phase difference between voltage and current equals:
\[Z =199.8 \;\;Ω\;\;\;\;\;\;-C\]
Voltage and current are in phase
\[Z =250.4 \;\;Ω \;\;\;\;\;\;-A\]
Current leads voltage by (𝜋/2)
\[ Z =153.7 \;\;Ω\;\;\;\;\;-D\]
Current leads voltage by (𝜋/3)
\[Z =250.4 \;\;Ω \;\;\;\;\;\;-B\]
Voltage leads current by (𝜋/2)
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\[14\star\]
An AC circuit containing a pure inductive coil had its source frequency changed,
and a graph was plotted between frequency and coil reactance resulting in the following graph.
The self-inductance coefficient of the coil equals:
\[L = 0.6 H\;\;\;\;\;\;-C\]
\[L = 0.2 H \;\;\;\;\;\;-A\]
\[ L = 0.8 H\;\;\;\;\;-D\]
\[ L = 0.4 H \;\;\;\;\;\;-B\]
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\[15\star\]
An AC circuit contains a capacitive capacitor. The frequency of the source was changed and a graph was plotted between the inverse of the frequency and the capacitive reactance, resulting in the following graph. The capacitance of the capacitor is equivalent to:
\[ C = 4\;\; 𝜇F \;\;\;\;\;\;-C\]
\[ C = 2\;\; 𝜇F \;\;\;\;\;\;-A\]
\[ C = 5 \;\;𝜇F\;\;\;\;\;-D\]
\[ C = 3\;\; 𝜇F \;\;\;\;\;\;-B\]
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\[16\star\]
An oscillation circuit consisting of a coil and a capacitor was charged in some way
and left to oscillate. When the current becomes zero in
the coil, then:
The capacitor is charged and the -C
electrical energy is zero and the magnetic energy is zero
The capacitor is charged and -A
the electrical energy is at maximum and the magnetic energy is zero
The capacitor is not charged and -D
the electrical energy is zero and the magnetic energy is zero
The capacitor is not charged and -B
the electrical energy is zero and the magnetic energy is at maximum
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\[17\star\]
A circuit containing a coil, resistor and capacitor
\[RLC\] connected to an AC source with maximum voltage \[V_m=240V \] and the maximum voltage across the resistor \[V_{mR}=120 V\]. The frequency of the source was changed until the maximum voltage across the coil equals the maximum voltage across the capacitor
\[V{mL} = V_{mC } \]
Then the maximum current in the circuit is:
Given that \[R=160 Ω\]
\[ I_m = 2 \;\;A \;\;\;\;\;\;-C\]
\[ I_m = 3 \;\;A \;\;\;\;\;\;-A\]
\[ I_m = 0.5 \;\;A \;\;\;\;\;-D\]
\[ I_m = 1.5 \;\;A \;\;\;\;\;\;-B\]
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\[18\star\]
A rectangular circuit
\[RLC\] connected to an antenna contains a coil with self-inductance
\[L=0.03 H\]
and a capacitor with capacity
\[C=2𝜇𝐹\] then it
is capable of receiving a wave with frequency
\[ 𝑓= 425 \;\;HZ \;\;\;\;\;\;-C\]
\[𝑓= 320 ;\;HZ \;\;\;\;\;\;-A\]
\[ 𝑓= 520\;\; HZ\;\;\;\;\;-D\]
\[ 𝑓= 650\;\; HZ\;\;\;\;\;\;-B\]
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\[19\star\]
For an
\[RLC\] circuit, the relationship between circuit impedance and frequency
was plotted and the following graph was obtained. Given that
\[ C= 5 nF\], the self-inductance coefficient of the coil is
\[L = 0.56\;\; H\;\;\;\;\;\;-C\]
\[L = 0.34\;\; H \;\;\;\;\;\;-A\]
\[ L = 0.24 \;\;H\;\;\;\;\;-D\]
\[ L = 0.73\;\; H \;\;\;\;\;\;-B\]
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\[20\star\]
In the following circuits, the phase angle between current and voltage is the largest possible, given that in any circuit always \[XC>XL\]
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\[21\star\]
Three lamps are connected as shown in the figure to an AC source and have the same brightness. When the source frequency is increased, one of the following occurs:
The brightness of the lamp connected - C
to the resistor increases
The brightness of the lamp connected - A
to the capacitor increases
All lamps' brightness remains unchanged - D
The brightness of the lamp connected - B
to the inductor increases
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\[22\star\]
An AC circuit contains a capacitor with capacitance \[C=4.4μF\] and an ohmic resistor as shown in the figure. They are connected to an AC source with frequency \[f= 60 HZ\]. If the total impedance of the circuit is \[Z=1000 Ω \], then the value of the resistance equals
\[ R= 797.9 \;\;Ω\;\;\;\;\;\;-C\]
\[R= 388.1\;\; Ω \;\;\;\;\;\;-A\]
\[ R= 654.3\;\; Ω \;\;\;\;\;-D\]
\[ R= 244.2\;\;Ω \;\;\;\;\;\;-B\]
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\[23\star\]
The following graph shows the effective changes in current intensity with the change in self-inductance coefficient of a coil in the circuit shown below. The frequency of the source equals
\[ 𝑓= 262.8\;\;HZ \;\;\;\;\;\;-C\]
\[𝑓= 480.4 ;\;HZ \;\;\;\;\;\;-A\]
\[ 𝑓= 100.2\;\; HZ\;\;\;\;\;-D\]
\[ 𝑓= 168.3\;\; HZ\;\;\;\;\;\;-B\]
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\[24\star\]
A step-down transformer with winding ratio
\[12:6\]
is connected to an AC source with effective voltage \[110 V\] and the other end is connected to a lamp. The effective voltage that reaches the lamp equals
\[ V_e= 55\;\; V\;\;\;\;\;\;-C\]
\[V_e= 80\;\; V \;\;\;\;\;\;-A\]
\[ V_e= 12\;\; V\;\;\;\;\;-D\]
\[ V_e= 220\;\; V\;\;\;\;\;\;-B\]
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25
An ideal transformer has fewer turns in its primary coil than in the secondary coil. It is connected to an AC power source and the other end is connected to a device. The characteristics of this transformer are:
Current step-up, voltage step-down -C
Current step-up, voltage step-up -A
Current step-down, voltage step-up -D
Current step-down, voltage step-down -B
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\[26\star\]
An ideal mobile phone charger converts electrical voltage
\[230 V\Rightarrow 5V\]
(3A) The phone needs current
when charging. The effective current required from the source equals
\[ IP = 0.025 \;\;A \;\;\;\;\;\;-C\]
\[ IP = 0.065 \;\;A \;\;\;\;\;\;-A\]
\[ IP = 0.04 \;\;A \;\;\;\;\;\;-D\]
\[ IP = 0.015 \;\;A \;\;\;\;\;\;-B\]
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\[27\star\]
In the circuit shown below, the voltmeter reading for the resistor equals
\[VR=150\;\; V \;\;\;\;\;\;-C\]
\[ VR=300\;\; V \;\;\;\;\;\;-A\]
\[VR=100\;\; V\;\;\;\;\;\;-D\]
\[VR=200 \;\;V \;\;\;\;\;\;-B\]
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\[28\star\]
An inductive coil and a lamp were connected to a DC source
then to an AC source and an iron core was inserted
left inside the coil
as shown in the figure below
What happens to the brightness of the lamp in both circuits
In figure 1 the lamp brightness decreases -C
and returns, and in figure 2 it decreases and does not return
In both cases the lamp brightness -A
increases and does not return
In figure 2 the lamp brightness decreases -D
and returns, and in figure 1 it decreases and does not return
In both cases the lamp brightness -B
increases and does not return
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\[29\star\]
One of the following circuits could have zero impedance
in the circuit
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\[30\star\]
In an
\[RLC\] circuit when in resonance state
the reading of one of the voltmeters shown
in the figure below equals zero
\[ V3=0 \;\;\;\;\;\;-C\]
\[ V1=0 \;\;\;\;\;\;-A\]
\[ V4=0 \;\;\;\;\;-D\]
\[ V2=0 \;\;\;\;\;\;-B\]
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\[31\star\]
In transformers, the iron core is made of insulated sheets separated from each other and the reason for this is
Reduce loss from field lines -C
Prevent eddy currents -A
To form mutual induction between the two coils -D
Reduce heat loss -B
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\[32\star\]
A circuit with a resistor and a capacitor
\[RC\] connected to an AC source with equation \[V (t)=210 Sin (200 t)\]
If the capacitor capacity is
\[ C= 50 𝜇F \]
and the equation of voltage difference across the capacitor is \[V (t)=150 Sin (200 t)\] then the resistance value equals
A series RLC circuit is connected to an AC source. If the capacitive reactance is less than the inductive reactance \[XL > XC \], then one of the following statements is correct:
(θ ) Current leads voltage by -C |
(θ) Voltage leads current by -A |
(𝜋/2 ) Voltage leads current by -D |
Current and voltage are in phase -B |
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In an LC circuit, a capacitor is charged by a battery and then connected to a pure inductor. When the current in the circuit is at its maximum value, then:
Electrical energy is at maximum -C |
Capacitor is fully charged -A |
Magnetic energy is at maximum -D |
Inductor stores energy and capacitor -B
|
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In an oscillating circuit, a capacitor with capacitance \[6 n𝐹\] and charge \[3 𝜇c\] is connected to an inductor with self-inductance coefficient \[0.2 H \]. One of the following values cannot be the current intensity in the circuit:
\[ i = - 0.042\;\; A \;\;\;\;\;\;-C\] |
\[ i = 0.092\;\; A\;\;\;\;\;\;-A\] |
\[ i = - 0.02 \;\;A \;\;\;\;\;-D\] |
\[ i = 0.082 \;\;A\;\;\;\;\;\;-B\] |
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In an LC circuit, during one cycle, the electrical energy equals the magnetic energy
\[\frac{1}{2}\frac{q^2}{c}=\frac{1}{2}L.i^2\]
Three times -C |
Once -A |
Four times -D |
Twice -B |
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In an AC circuit containing a single element (parallel plate capacitor - cylindrical capacitor - inductor - ohmic resistor) the source frequency was increased and it was observed that the maximum value of the alternating current did not change. Therefore, the circuit contains:
Inductor -C |
Parallel plate capacitor -A |
Ohmic resistor -D |
Cylindrical capacitor -B |
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At what frequency does the reactance of an inductor with self-inductance coefficient \[0.4 H \] reach an impedance of \[93 Ω\]?
\[ f = 37\;\; HZ\;\;\;\;\;\;-C\] |
\[ f = 35\;\; HZ \;\;\;\;\;\;-A\] |
\[f = 38 \;\;HZ \;\;\;\;\;-D\] |
\[ f = 36 \;\;HZ\;\;\;\;\;\;-B\] |
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A capacitive capacitor with capacitance
C= 4.0 × 10–4F
is connected to an AC source
with source voltage equation of the form
\[ V (t) = 110 sin (126t )\]
The maximum current passing through the capacitor equals:
\[ i_{max}=1.42 \;\;A\;\;\;\;\;\;-C\] |
\[ i_{max}=5.54 \;\;A \;\;\;\;\;\;-A\] |
\[ i_{max}=3.85 \;\;A\;\;\;\;\;-D\] |
\[ i_{max}=2.63 \;\;A\;\;\;\;\;\;-B\] |
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In an AC circuit, the ammeter and voltmeter were monitored
and it was observed that the current lags behind the voltage in phase by
\[𝜑=\frac{𝜋}{2}=90^0\]. Therefore, the circuit contains:
Capacitive capacitor -C |
Ohmic resistor -A |
None of the above devices -D |
Pure inductor -B |
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In the adjacent figure, an
\[RLC\]
circuit is connected to an AC source with equation \[V(t) = 210 Sin (300 t ) \]. If the self-inductance coefficient of the inductor,
the capacitance of the capacitor, and the ohmic resistance are
\[ L= 0.5 H \;\;\;\;\;\;\;\;C= 5 𝜇F\;\;\;\;\;\;\;\;R=150 Ω\]
then the impedance of the circuit equals:
\[ Z= 335\;\; Ω \;\;\;\;\;\;-C\] |
\[ Z= 620\;\; Ω \;\;\;\;\;\;-A\] |
\[ Z= 427\;\; Ω\;\;\;\;\;-D\] |
\[ Z= 538\;\; Ω \;\;\;\;\;\;-B\] |
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In the adjacent figure, an
\[RLC\]
circuit is shown. If the maximum voltage
of the source is \[V_m = 200 V \], the maximum voltage across the resistor is \[V_mR = 150 V \], and the maximum voltage
across the inductor is \[V_mL = 180 V \], then the maximum voltage across the capacitor equals:
\[ VC =120 \;\;V\;\;\;\;\;\;-C\] |
\[ VC =47.7 \;\;V \;\;\;\;\;\;-A\] |
\[ VC =253.2\;\; V \;\;\;\;\;-D\] |
\[ VC =93.2\;\; V \;\;\;\;\;\;-B\] |
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An AC circuit connected to a time-varying EMF source with the equation \[V(t) = 210 Sin (60 𝜋 t ) \] is connected to an inductor with self-inductance \[L=0.2 H \]. The equation of the alternating current representing this circuit is:
\[ 𝐼(t) = 5.57 Sin (60𝜋 t - 𝜋/2 )\;\;\;\;\;\;-C\] |
\[ 𝐼(t) = 5.67 Sin (60𝜋 t + 𝜋/2 ) \;\;\;\;\;\;-A\] |
\[ 𝐼(t) = 17.5 Sin (60𝜋 t - 𝜋/2 )\;\;\;\;\;-D\] |
\[ 𝐼(t) = 17.5 Sin (60𝜋 t + 𝜋/2 ) \;\;\;\;\;\;-B\] |
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An AC circuit is connected to a time-varying EMF source. The source voltage equation is \[V(t) = 210 Sin (60 𝜋 t )\]. The source is connected to an ohmic resistor \[R=20 Ω\]. The AC current equation representing this circuit is:
\[ 𝐼(t) = 10.5 Sin (60𝜋 t )\;\;\;\;\;\;-C\] |
\[ 𝐼(t) = 10.5 Sin (60𝜋 t+ 𝜋/2 ) \;\;\;\;\;\;-A\] |
\[ 𝐼(t) = 14.7 Sin (60𝜋 t +𝜋/2)\;\;\;\;\;-D\] |
\[ 𝐼(t) = 14.7 Sin (60𝜋 t -𝜋/2) \;\;\;\;\;\;-B\] |
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In the adjacent figure, an LC circuit is connected to an AC source with equation \[V(t) = 110 Sin (300 t )\]. If the self-inductance coefficient of the coil is \[L= 0.4 H\] and the capacitor capacity is \[ C= 9 𝜇F \], then the impedance value of the circuit and the phase difference between voltage and current equals:
\[Z =199.8 \;\;Ω\;\;\;\;\;\;-C\] Voltage and current are in phase |
\[Z =250.4 \;\;Ω \;\;\;\;\;\;-A\] Current leads voltage by (𝜋/2) |
\[ Z =153.7 \;\;Ω\;\;\;\;\;-D\] Current leads voltage by (𝜋/3) |
\[Z =250.4 \;\;Ω \;\;\;\;\;\;-B\] Voltage leads current by (𝜋/2) |
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An AC circuit containing a pure inductive coil had its source frequency changed, and a graph was plotted between frequency and coil reactance resulting in the following graph. The self-inductance coefficient of the coil equals:
\[L = 0.6 H\;\;\;\;\;\;-C\] |
\[L = 0.2 H \;\;\;\;\;\;-A\] |
\[ L = 0.8 H\;\;\;\;\;-D\] |
\[ L = 0.4 H \;\;\;\;\;\;-B\] |
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An AC circuit contains a capacitive capacitor. The frequency of the source was changed and a graph was plotted between the inverse of the frequency and the capacitive reactance, resulting in the following graph. The capacitance of the capacitor is equivalent to:
\[ C = 4\;\; 𝜇F \;\;\;\;\;\;-C\] |
\[ C = 2\;\; 𝜇F \;\;\;\;\;\;-A\] |
\[ C = 5 \;\;𝜇F\;\;\;\;\;-D\] |
\[ C = 3\;\; 𝜇F \;\;\;\;\;\;-B\] |
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An oscillation circuit consisting of a coil and a capacitor was charged in some way
and left to oscillate. When the current becomes zero in
the coil, then:
The capacitor is charged and the -C
|
The capacitor is charged and -A
|
The capacitor is not charged and -D
|
The capacitor is not charged and -B
|
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A circuit containing a coil, resistor and capacitor
\[RLC\] connected to an AC source with maximum voltage \[V_m=240V \] and the maximum voltage across the resistor \[V_{mR}=120 V\]. The frequency of the source was changed until the maximum voltage across the coil equals the maximum voltage across the capacitor
\[V{mL} = V_{mC } \]
Then the maximum current in the circuit is:
Given that \[R=160 Ω\]
\[ I_m = 2 \;\;A \;\;\;\;\;\;-C\] |
\[ I_m = 3 \;\;A \;\;\;\;\;\;-A\] |
\[ I_m = 0.5 \;\;A \;\;\;\;\;-D\] |
\[ I_m = 1.5 \;\;A \;\;\;\;\;\;-B\] |
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A rectangular circuit \[RLC\] connected to an antenna contains a coil with self-inductance \[L=0.03 H\] and a capacitor with capacity \[C=2𝜇𝐹\] then it is capable of receiving a wave with frequency
\[ 𝑓= 425 \;\;HZ \;\;\;\;\;\;-C\] |
\[𝑓= 320 ;\;HZ \;\;\;\;\;\;-A\] |
\[ 𝑓= 520\;\; HZ\;\;\;\;\;-D\] |
\[ 𝑓= 650\;\; HZ\;\;\;\;\;\;-B\] |
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For an
\[RLC\] circuit, the relationship between circuit impedance and frequency
was plotted and the following graph was obtained. Given that
\[ C= 5 nF\], the self-inductance coefficient of the coil is
\[L = 0.56\;\; H\;\;\;\;\;\;-C\] |
\[L = 0.34\;\; H \;\;\;\;\;\;-A\] |
\[ L = 0.24 \;\;H\;\;\;\;\;-D\] |
\[ L = 0.73\;\; H \;\;\;\;\;\;-B\] |
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In the following circuits, the phase angle between current and voltage is the largest possible, given that in any circuit always \[XC>XL\]
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Three lamps are connected as shown in the figure to an AC source and have the same brightness. When the source frequency is increased, one of the following occurs:
|
The brightness of the lamp connected - C
|
The brightness of the lamp connected - A
|
|
All lamps' brightness remains unchanged - D |
The brightness of the lamp connected - B
|
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An AC circuit contains a capacitor with capacitance \[C=4.4μF\] and an ohmic resistor as shown in the figure. They are connected to an AC source with frequency \[f= 60 HZ\]. If the total impedance of the circuit is \[Z=1000 Ω \], then the value of the resistance equals
\[ R= 797.9 \;\;Ω\;\;\;\;\;\;-C\] |
\[R= 388.1\;\; Ω \;\;\;\;\;\;-A\] |
\[ R= 654.3\;\; Ω \;\;\;\;\;-D\] |
\[ R= 244.2\;\;Ω \;\;\;\;\;\;-B\] |
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The following graph shows the effective changes in current intensity with the change in self-inductance coefficient of a coil in the circuit shown below. The frequency of the source equals
\[ 𝑓= 262.8\;\;HZ \;\;\;\;\;\;-C\] |
\[𝑓= 480.4 ;\;HZ \;\;\;\;\;\;-A\] |
\[ 𝑓= 100.2\;\; HZ\;\;\;\;\;-D\] |
\[ 𝑓= 168.3\;\; HZ\;\;\;\;\;\;-B\] |
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A step-down transformer with winding ratio \[12:6\] is connected to an AC source with effective voltage \[110 V\] and the other end is connected to a lamp. The effective voltage that reaches the lamp equals
\[ V_e= 55\;\; V\;\;\;\;\;\;-C\] |
\[V_e= 80\;\; V \;\;\;\;\;\;-A\] |
\[ V_e= 12\;\; V\;\;\;\;\;-D\] |
\[ V_e= 220\;\; V\;\;\;\;\;\;-B\] |
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An ideal transformer has fewer turns in its primary coil than in the secondary coil. It is connected to an AC power source and the other end is connected to a device. The characteristics of this transformer are:
Current step-up, voltage step-down -C |
Current step-up, voltage step-up -A |
Current step-down, voltage step-up -D |
Current step-down, voltage step-down -B |
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An ideal mobile phone charger converts electrical voltage
\[230 V\Rightarrow 5V\]
(3A) The phone needs current
when charging. The effective current required from the source equals
\[ IP = 0.025 \;\;A \;\;\;\;\;\;-C\] |
\[ IP = 0.065 \;\;A \;\;\;\;\;\;-A\] |
\[ IP = 0.04 \;\;A \;\;\;\;\;\;-D\] |
\[ IP = 0.015 \;\;A \;\;\;\;\;\;-B\] |
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In the circuit shown below, the voltmeter reading for the resistor equals
\[VR=150\;\; V \;\;\;\;\;\;-C\] |
\[ VR=300\;\; V \;\;\;\;\;\;-A\] |
\[VR=100\;\; V\;\;\;\;\;\;-D\] |
\[VR=200 \;\;V \;\;\;\;\;\;-B\] |
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An inductive coil and a lamp were connected to a DC source
then to an AC source and an iron core was inserted
left inside the coil
as shown in the figure below
What happens to the brightness of the lamp in both circuits
In figure 1 the lamp brightness decreases -C
|
In both cases the lamp brightness -A
|
In figure 2 the lamp brightness decreases -D
|
In both cases the lamp brightness -B
|
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One of the following circuits could have zero impedance
in the circuit
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In an
\[RLC\] circuit when in resonance state
the reading of one of the voltmeters shown
in the figure below equals zero
\[ V3=0 \;\;\;\;\;\;-C\] |
\[ V1=0 \;\;\;\;\;\;-A\] |
\[ V4=0 \;\;\;\;\;-D\] |
\[ V2=0 \;\;\;\;\;\;-B\] |
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In transformers, the iron core is made of insulated sheets separated from each other and the reason for this is
Reduce loss from field lines -C |
Prevent eddy currents -A |
To form mutual induction between the two coils -D |
Reduce heat loss -B |
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A circuit with a resistor and a capacitor
\[RC\] connected to an AC source with equation \[V (t)=210 Sin (200 t)\]
If the capacitor capacity is
\[ C= 50 𝜇F \]
and the equation of voltage difference across the capacitor is \[V (t)=150 Sin (200 t)\] then the resistance value equals