Question 1

The ideal meters shown read rms current and voltage.The average power delivered to the load is: 
A)definitely not equal to VI
B)perhaps more than VI
C)possibly equal to VI even if the load contains an inductor and a capacitor
D)definitely less than VI
E)zero as is the average of any sine wave

Accepted Answer

The answer of The ideal meters shown read rms current...

Question 2

In order to maximize the rate at which energy is supplied to a resistive load, the power factor of an RLC circuit should be as close as possible to:
A)0
B)0.5
C)1
D)infinity
E)cannot tell without knowing R, L, C, and the driving frequency

Accepted Answer

The power factor of an RLC circuit is defined as the cosine of the phase angle between the voltage and current waveforms. A power factor of 1 (cosine of 0 degrees) indicates that the voltage and current are perfectly in phase, which means that all of the energy delivered to the circuit is being used by the load. Therefore, a power factor as close to 1 as possible will maximize the rate at which energy is supplied to the resistive load.

Question 3

The rms value of an ac current is:
A)its peak value
B)its average value
C)that steady current that produces the same rate of heating in a resistor
D)that steady current that will charge a battery at the same rate
E)zero

Accepted Answer

The rms (root mean square) value of an AC current is defined as that steady current which, when passed through a given resistor for a given time, produces the same amount of heat as the AC current does in the same resistor over the same period. This is a measure of the effective value of the varying current.

Question 4

For a power transmission line, the transmission should be at low current and high voltage because:
A)this is the least dangerous to electrical workers
B)this gives the least dose of electromagnetic radiation to the public
C)this minimizes transmission losses
D)low current and high voltage is easier to transform than high current and low voltage
E)household appliances run at low current and high voltage

Accepted Answer

Low current and high voltage transmission reduces the resistance losses in the transmission lines, which leads to minimized transmission losses. This is the main reason why power transmission lines use high voltage and low current.

Question 5

An RLC circuit has a sinusoidal source of emf.The average rate at which the source supplies energy is 5 nW.This must also be:
A)the average rate at which energy is stored in the capacitor
B)the average rate at which energy is stored in the inductor
C)the average rate at which energy is dissipated in the resistor
D)twice the average rate at which energy is stored in the capacitor
E)three times the average rate at which energy is stored in the inductor

Accepted Answer

In an RLC circuit, the energy supplied by the source is dissipated as heat in the resistor. Therefore, the average rate at which energy is dissipated in the resistor is equal to the average rate at which the source supplies energy. The average rate at which energy is stored in the capacitor and the inductor can be calculated using the formula for energy stored in a capacitor and inductor respectively, but these values may not necessarily be equal to the average rate at which the source supplies energy. Therefore, choices A, B, D, and E are incorrect.

Question 6

In a sinusoidally driven series RLC circuit, the inductive resistance is X_L = 200 $\Omega$, the capacitive reactance is X_C = 100 $\Omega$, and the resistance is R = 50 $\Omega$.The current and applied emf would be in phase if: A)the resistance is increased to 100 $\Omega$, with no other changes B)the resistance is increased to 200 $\Omega$, with no other changes C)the inductance is reduced to zero, with no other changes D)the capacitance is doubled, with no other changes E)the capacitance is halved, with no other changes

Accepted Answer

The answer of In a sinusoidally driven series RLC circuit,...

Question 7

A power transmission line carries 400A of current at a voltage of 765 kV.If the line has a resistance of 29 µΩ/m, what is the rate at which energy is being dissipated in 800 km of line?
A)0 W
B)3.7 kW
C)310 kW
D)3.7 MW
E)310 MW

Accepted Answer

First, we use the formula for power dissipation in a resistance:
Power = (Current^2) x Resistance
We need to find the total resistance of the line, which can be calculated by multiplying the resistance per unit length by the length of the line:
Total resistance = resistance per unit length x length of line = 29 µΩ/m x 800,000 m = 23,200 µΩ

Next, we can calculate the power dissipated in the entire line:
Power = (400^2) x 23,200 µΩ
Power = 3,728,000 µW

Converting to megawatts:
Power = 3.728 MW

Therefore, the rate at which energy is being dissipated in 800 km of line is 3.7 MW. Answer: D.

Question 8

In a series RLC circuit the rms value of the generator emf is and the rms value of the current is i.The current lags the emf by $\phi$.The average power supplied by the generator is given by: A)(i/2)cos $\phi$ B)i C)i²/Z D)i²Z E)i²R

Accepted Answer

The average power supplied by the generator in an AC circuit is given by P = I^2R, where I is the rms current and R is the resistance.

Question 9

A series circuit consists of a 15-$\Omega$ resistor, a 25-mH inductor, and a 35-$\mu$F capacitor.If the frequency is 100 Hz the power factor is:
A)0
B)0.20
C)0.45
D)0.65
E)1.0

Accepted Answer

To find the power factor, we need to first calculate the impedance Z of the circuit.

Impedance Z = sqrt(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

R = 15 ohms
XL = 2*pi*f*L = 2*pi*100*0.025 = 15.7 ohms
XC = 1/(2*pi*f*C) = 1/(2*pi*100*35*10^-6) = 45.5 ohms

Therefore, Z = sqrt(15^2 + (15.7 - 45.5)^2) = 45.6 ohms

The power factor (PF) = cos(theta) = R/Z = 15/45.6 = 0.328

So, the closest option is C, which is 0.45.

Question 10

An RLC series circuit, connected to a source , is at resonance.Then:
A)the voltage across R is zero
B)the voltage across R equals the applied voltage
C)the voltage across C is zero
D)the voltage across L equals the applied voltage
E)the applied voltage and current differ in phase by 90  $\degree$

Accepted Answer

At resonance in an RLC series circuit, the impedance is purely resistive, so the voltage across the resistor R equals the applied voltage.

Question 11

The main reason that alternating current replaced direct current for general use is:
A)ac generators do not need slip rings
B)ac voltages may be conveniently transformed
C)electric clocks do not work on dc
D)a given ac current does not heat a power line as much as the same dc current
E)ac minimizes magnetic effects

Accepted Answer

AC voltages can be transformed using transformers, which allows for more efficient transmission over long distances and the ability to adjust the voltage level for different applications. This was a major advantage over DC, which could not easily be transformed.

Question 12

The units of the power factor are: A)ohm B)watt C)radian D)ohm^1/2 E)none of these

Accepted Answer

The power factor is a dimensionless number (having no unit) that represents the cosine of the phase angle between voltage and current in an AC circuit, indicating the ratio of real power flowing to the load to the apparent power in the circuit.

Question 13

A sinusoidal voltage V(t)has an rms value of 100 V.Its maximum value is:
A)70.7 V
B)100 V
C)141 V
D)200 V
E)707 V

Accepted Answer

The rms value of a sinusoidal voltage is equal to its maximum (or peak) value divided by the square root of 2 (i.e. Vrms = Vmax / √2). Rearranging this equation, we can solve for Vmax: Vmax = Vrms x √2. Plugging in the given value for Vrms (100 V), we get Vmax = 100 V x √2 ≈ 141 V. Therefore, the correct answer is choice C.

Question 14

The core of a transformer is made in a laminated form to: A)facilitate easy assembly B)reduce i²R losses in the coils C)increase the magnetic flux D)save weight E)prevent eddy currents

Accepted Answer

Laminating the core of a transformer helps in reducing eddy currents, which are loops of electrical current induced within the core by the changing magnetic field. This, in turn, reduces energy losses and improves efficiency.

Question 15

In a sinusoidally driven series RLC circuit the current lags the applied emf.The rate at which energy is dissipated in the resistor can be increased by:
A)decreasing the capacitance and making no other changes
B)increasing the capacitance and making no other changes
C)increasing the inductance and making no other changes
D)increasing the driving frequency and making no other changes
E)decreasing the amplitude of the driving emf and making no other changes

Accepted Answer

The answer of In a sinusoidally driven series RLC circuit...

Question 16

An RLC series circuit is driven by a sinusoidal emf with angular frequency $ \omega $_d.If $ \omega $_d is increased without changing the amplitude of the emf, the current amplitude increases.If L is the inductance, C is the capacitance, and R is the resistance, this means that: A)" $ \omega $_dL > 1/ $ \omega $_dC" B)" $ \omega $_dL < 1/ $ \omega $_dC" C)" $ \omega $_dL = 1/ $ \omega $_dC" D)" $ \omega $_dL > R" E)" $ \omega $_dL < R"

Accepted Answer

The resonant frequency of an RLC circuit is given by f = 1/(2π√(LC)). As ω = 2πf, we can rewrite the resonant frequency as ω = 1/√(LC). When ω is increased, the denominator decreases, leading to an increase in the resonant frequency. For an RLC circuit with fixed values of L and C, increasing ω will result in the resonant frequency being greater than the driving frequency, causing the current amplitude to increase. This happens only when the inductive reactance is greater than the capacitive reactance. Therefore, we must have: ωL > 1/ωC or L/C < 1/ ω^2. Rewriting in terms of resonant frequency: L/C < 1/(2πf)^2 or LC > (2πf)^2. Hence, the correct answer is B: "F1F1F1S1ωF1F1F-1_dL < 1/F1F1F1S1ωF1F1F-1_dC."

Question 17

Iron, rather than copper, is used in the core of transformers because iron:
A)can withstand a higher temperature
B)has a greater resistivity
C)has a very high permeability
D)makes a good permanent magnet
E)insulates the primary from the secondary

Accepted Answer

Iron has a very high permeability, meaning that it can more efficiently conduct and transfer magnetic fields. This makes it a better choice for the core of transformers, where magnetic fields need to be transferred from the primary coil to the secondary coil.

Question 18

The average power supplied to the circuit shown passes through a maximum when which one of the following is increased continuously from a very low to a very high value? 
A)source emf
B)R
C)C
D)source frequency f
E)none of these

Accepted Answer

The answer of The average power supplied to the circuit...

Question 19

A generator supplies 100 V to the primary coil of a transformer.The primary has 50 turns and the secondary has 500 turns.The secondary voltage is:
A)1000 V
B)500 V
C)250 V
D)100 V
E)10 V

Accepted Answer

The secondary voltage can be calculated using the transformer equation: $V_s = V_p \times \frac{N_s}{N_p}$, where $V_s$ is the secondary voltage, $V_p$ is the primary voltage, $N_s$ is the number of turns in the secondary coil, and $N_p$ is the number of turns in the primary coil. Substituting the given values: $V_s = 100 V \times \frac{500}{50} = 1000 V$.

Question 20

An RC series circuit is connected to an emf source having angular frequency $ \omega $.The current: A)leads the source emf by tan^-1(1/ $ \omega $CR) B)lags the source emf by tan^-1(1/ $ \omega $CR) C)leads the source emf by tan^-1( $ \omega $CR) D)lags the source emf by tan^-1( $ \omega $CR) E)leads the source emf by $ \omega $/4

Accepted Answer

The answer of An RC series circuit is connected to...

Quiz 31: Electromagnetic Oscillations and Alternating Current

The ideal meters shown read rms current and voltage.The average power delivered to the load is:

In order to maximize the rate at which energy is supplied to a resistive load, the power factor of an RLC circuit should be as close as possible to:

The rms value of an ac current is:

For a power transmission line, the transmission should be at low current and high voltage because:

An RLC circuit has a sinusoidal source of emf.The average rate at which the source supplies energy is 5 nW.This must also be:

In a sinusoidally driven series RLC circuit, the inductive resistance is X_L = 200 $\Omega$ , the capacitive reactance is X_C = 100 $\Omega$ , and the resistance is R = 50 $\Omega$ .The current and applied emf would be in phase if:

A power transmission line carries 400A of current at a voltage of 765 kV.If the line has a resistance of 29 µΩ/m, what is the rate at which energy is being dissipated in 800 km of line?

In a series RLC circuit the rms value of the generator emf is and the rms value of the current is i.The current lags the emf by $\phi$ .The average power supplied by the generator is given by:

A series circuit consists of a 15- $\Omega$ resistor, a 25-mH inductor, and a 35- $\mu$ F capacitor.If the frequency is 100 Hz the power factor is:

An RLC series circuit, connected to a source , is at resonance.Then:

The main reason that alternating current replaced direct current for general use is:

The units of the power factor are:

A sinusoidal voltage V(t) has an rms value of 100 V.Its maximum value is:

The core of a transformer is made in a laminated form to:

In a sinusoidally driven series RLC circuit the current lags the applied emf.The rate at which energy is dissipated in the resistor can be increased by:

An RLC series circuit is driven by a sinusoidal emf with angular frequency $\omega$ _d.If $\omega$ _d is increased without changing the amplitude of the emf, the current amplitude increases.If L is the inductance, C is the capacitance, and R is the resistance, this means that:

Iron, rather than copper, is used in the core of transformers because iron:

The average power supplied to the circuit shown passes through a maximum when which one of the following is increased continuously from a very low to a very high value?

A generator supplies 100 V to the primary coil of a transformer.The primary has 50 turns and the secondary has 500 turns.The secondary voltage is:

An RC series circuit is connected to an emf source having angular frequency $\omega$ .The current: