Electromagnetic Oscillations and Alternating Current MCQ Quiz - Objective Question with Answer for Electromagnetic Oscillations and Alternating Current - Download Free PDF
Last updated on May 21, 2025
Latest Electromagnetic Oscillations and Alternating Current MCQ Objective Questions
Electromagnetic Oscillations and Alternating Current Question 1:
To an AC power supply of 220 V at 50 Hz, a resistor of 20 Ω, a capacitor of reactance 25 Ω and an inductor of reactance 45 Ω are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 1 Detailed Solution
Correct option is: (2) 7.8 A and 45°
XL = 45 W, XC = 25 W, R = 20 W
⇒ I = 220 / √((XL − XC)2 + R2) = 220 / √((45 − 25)2 + 202)
⇒ = 220 / (2√2) = 11 / √2 = 7.779 A
⇒ tan φ = (XL − XC) / R = (45 − 25) / 20 = 1
⇒ φ = 45°
Electromagnetic Oscillations and Alternating Current Question 2:
These are connected to variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2 (indicated in circuits) are related as shown in Column I.
Column I | Column II |
---|---|
(A) I ≠ 0, V1 ∝ I | (p) |
(B) I ≠ 0, V2 > V1 | (q) |
(C) V1 = 0, V2 = V | (r) |
(D) I ≠ 0, V2 ∝ I | (s) |
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 2 Detailed Solution
Calculation:
In circuit (p): Under steady state, the capacitor will act as an infinite impedance and the inductor will act as a zero impedance. Thus:
I = 0
V1 = 0
V2 = V
In circuit (q): The inductor will act as zero resistance in steady state, giving us:
I = V / R = V / 2
V1 = 0
V2 = V
In circuit (r): The inductive reactance XL and total impedance Z are:
XL = ωL = 2πνL = 1.88 Ω
Z = √(XL2 + R2) = 2.75 Ω
Thus,
I = V / Z ≠ 0
V1 = XLI = 1.88I
V2 = RI = 2I
V2 > V1
In circuit (s): Inductive reactance XL, capacitive reactance XC, and impedance Z are:
XL = 1.88 Ω
XC = 1 / (ωC) = 1061 Ω
Z = XC − XL = 1059 Ω
So the current is:
I = V / Z ≠ 0
V1 = XLI = 1.88I
V2 = XCI = 1061I
V2 > V1
Answer: A → (r, s), B → (q, r, s), C → (p, q), D → (q, r, s )
Electromagnetic Oscillations and Alternating Current Question 3:
An \(AC\) voltage source of variable angular frequency \(\omega\) and fixed amplitude \(V_{0}\) is connected in series with a capacitance \(C\) and an electric bulb of resistance \(R\) (inductance zero). When \(\omega\) is increased
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 3 Detailed Solution
As \(\omega\) increases, \(Z\) decreases.
Hence, bulb will glow brighter.
Electromagnetic Oscillations and Alternating Current Question 4:
A resistor of 500 Ω, an inductance of 0.5 H are in series with an a. c. which is given by V = 100√2 sin (1000 t). The power factor of the combination is
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 4 Detailed Solution
CONCEPT:
- Voltage in AC: In an AC voltage source, the voltage of the source keeps changing with time and is defined as
V = V0 sin ωt
where V is the voltage at any time t, V0 is the max value of voltage, and ω is the angular frequency.
- With AC source capacitance, In a series combination of a resistor (R), an inductance (L)
Inductance Reactanceis defined as
\(X_L = ω L=2\pi fL\)
Where ω is the angular frequency, ƒ is the frequency in Hertz, C is the AC capacitance in Farads, and XL is the Inductive Reactance in Ohms,.
Overall Impedance (Z) in a series combination of a resistor (R), an inductance (L):
\(Z = \sqrt{R^2 + {ω L}^2}\)
where R is resistance and ωL is Inductance reactance.
- Power Factor: The power factor of series-connected L and R circuit in AC voltage is
cos ϕ = R/Z
where R is resistance and Z is overall impedance.
CALCULATION:
Given that L = 0.5H and R = 500 Ω
V = 100√2 sin (1000 t)
compare it with standard equation. V = V0 sin ωt
ω = 1000
\(X_L = ω L=1000 \times 0.5=500 \Omega\)
\(Z = \sqrt{R^2 + {ω L}^2} = \sqrt{500^2 + {500}^2}\)
\(Z = 500\sqrt2 \Omega\)
Power factor cos ϕ = R/Z
cos ϕ = \({500 \over500\sqrt2} ={ 1\over \sqrt2}\)
So the correct answer is option 1.
Electromagnetic Oscillations and Alternating Current Question 5:
In a series LCR circuit \(R=300\Omega, L=0.9H, C=2\mu F, \omega =1000\) rad/s. The impedance of the circuit is?
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 5 Detailed Solution
Calculation:
\(Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})}\)
\(=\sqrt{(300)^2+\left(1000\times 0.9-\dfrac{1}{1000\times 2\times 10^{-6}}\right)^2}\)
\(Z=\sqrt{(300)^2+(400)^2}\Rightarrow 500\Omega \)
The correct option is (a)
Top Electromagnetic Oscillations and Alternating Current MCQ Objective Questions
In Alternating Current (AC), the direction and magnitude of the current varies
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 6 Detailed Solution
Download Solution PDFThe correct answer is Periodically.
Key Points
CONCEPT:
- The electric current flows in two ways: Alternating current and Direct Current.
- Direct current flows only in one direction.
- Alternating current: The electric current whose direction changes periodically is called electric current.
- Alternating current reverses its direction periodically.
- It also changes its magnitude periodically because of the induced electromagnetic force.
- For Alternating current both magnitude and direction change. The frequency of the alternating current in the Indian power supply is 50 Hz. The time period is 1/50 = 20 msec.
EXPLANATION:
- In Alternating Current (AC), the direction and magnitude of the current vary periodically. So option 2 is correct.
Step-up transformers are used for
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 7 Detailed Solution
Download Solution PDFCONCEPT:
Transformer:
- An electrical device that is used to transfer electrical energy from one electrical circuit to another is called a transformer.
There are two types of transformer:
1. Step-up transformer:
- The transformer which increases the potential is called a step-up transformer.
- The number of turns in the secondary coil is more than that in the primary coil.
2.Step-down transformer:
- The transformer which decreases the potential is called a step-down transformer.
- The number of turns in the secondary coil is less than that in the primary coil.
EXPLANATION:
- As in the step-up transformer, the number of coils in the secondary coil is more than that in the primary coil. So the electrical potential increases.
- Hence step-up transformer is used for increasing the voltage/potential. So option 4 is correct.
- In step-up transformer, the current in the secondary coil is less than that in the primary coil.
The transformer core is laminated in order to
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 8 Detailed Solution
Download Solution PDFThe correct answer is Minimize eddy current loss.
- Transformer cores are laminated in order to minimize core loss.
- By providing laminations, the area of each part gets reduced and hence resistance will get very high which limits the eddy current to a minimum value, and hence eddy current losses get reduced.
- The laminations provide small gaps between the plates. As it is easier for magnetic flux to flow through iron than air or coil, the stray flux or leakage flux that can cause core losses is minimized.
The resonant frequency of a RLC circuit is equal to __________.
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 9 Detailed Solution
Download Solution PDFCONCEPT:
- The ac circuit containing the capacitor, resistor, and the inductor is called an LCR circuit.
- For a series LCR circuit, the total potential difference of the circuit is given by:
\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)
Where VR = potential difference across R, VL = potential difference across L and VC = potential difference across C
- For a series LCR circuit, Impedance (Z) of the circuit is given by:
\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive
CALCULATION:
- For a series LCR circuit, Impedance (Z) of the circuit is given by:
\(\)\(\Rightarrow Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
- Inductive reactance,
⇒ XL = Lω
- Capacitive reactance
\(\Rightarrow X_c=\frac{1}{C\omega}\)
- Resonance will take place when XL = XC.
⇒ XL = XC
\(\Rightarrow L\omega =\frac{1}{C\omega}\)
\(\Rightarrow \omega =\frac{1}{\sqrt{LC}}\)
A step-up transformer converts ________ into a ________.
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 10 Detailed Solution
Download Solution PDFCONCEPT:
Transformer:
An electrical device that is used to transfer electrical energy from one electrical circuit to another is called a transformer.
There are two types of transformer:
Step-up transformer:
- The transformer which increases the potential is called a step-up transformer.
- The number of turns in the secondary coil is more than that in the primary coil.
Step-down transformer:
- The transformer which decreases the potential is called a step-down transformer.
- The number of turns in the secondary coil is less than that in the primary coil.
EXPLANATION:
- As in the step-up transformer, the number of coils in the secondary coil is more than that in the primary coil. So the electrical potential increases.
- A step-up transformer converts low voltage at high current into a high voltage at low current. So option 4 is correct.
- In a step-up transformer, the current in the secondary coil is less than that in the primary coil.
- The secondary coil is made of thin insulated wire while the primary coil is made of thick insulated wire.
The peak value of an A.C. is 2√2 A, its rms value will be:
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 11 Detailed Solution
Download Solution PDFCONCEPT:
- Root mean square value of Alternating current (Irms): The value of steady current, which would generate the same amount of heat in a given resistance in a given time, as is done by the alternating current. when passed through the same resistance for the same time.
- The r.m.s. value is also called effective value or virtual value of alternating current.
The relation between the peak value of the alternating current value of current (Io) and r.m.s. value of the current is given as:
\({I_{rms}} = \frac{{{I_o}}}{{√ 2 }}\)
CALCULATION:
Given that:
Peak current (I0) = 2√2 A
\({I_{rms}} = \frac{{{I_o}}}{{√ 2 }}\)
\({I_{rms}} = \frac{{{2\sqrt2}}}{{√ 2 }} = 2 A\)
The rms value is 2 A.
An ac voltage v = vm sinωt applied to a capacitor drives a current in the capacitor, i = ______________.
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 12 Detailed Solution
Download Solution PDFConcept:
The power factor of the inductor has lagging nature, whereas for capacitor its leading.
For a pure inductor \(\phi = {90^0}\) (The angle between voltage and current)
This shows that if we plot voltage vs current graph the output signal will be lagging by phase angle π/2
The given figure shows the power factor (ϕ) for pure R, L and C circuit
Explanation:
From the above explanation, we can see that for a pure capacitance circuit if AC voltage of v = vm sinωt then the driven current in the capacitor will be
I = im sin(ωt + ϕ)
Now as mentioned for pure inductor phase angle will be lagging by 90° or π/2.
i.e., ϕ = π/2 ⇒ i = im sin (ωt + π/2)
In a step-up transformer, the value of current in the secondary coil in comparison to primary coil is
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 13 Detailed Solution
Download Solution PDFConcept:
- A transformer is an electrical device which transfers electrical energy from one circuit to another
- It works based on the principle of electromagnetic induction
- It is used for increasing or decreasing the amount of voltage or current as per our requirement based on these transformers are classified into two types step-up (increasing) or step-down (decreasing) transformer.
There are two types of transformer:
1. Step-up transformer:
- The transformer which increases the potential is called a step-up transformer.
- The number of turns in the secondary coil is more than that in the primary coil.
2.Step-down transformer:
- The transformer which decreases the potential is called a step-down transformer.
- The number of turns in the secondary coil is less than that in the primary coil.
Explanation:
As in an ideal transformer, there is no loss of power i.e.
Pout = Pin
So, VsIs = VpIp
\(\therefore \frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_P}}} = \frac{{{I_P}}}{{{I_S}}}\)
Where Ns = number of turns of the secondary coil, NP = number of turns of the primary coil, Vs = Voltage secondary coil, VP = Voltage primary coil, Is = Current in the secondary coil and IP = current in the primary coil.
In Step up transformer voltage is increased So, the current will be reduced. As there is no loss in power.
Additional Information
- A transformer can be used to either increase or decrease the voltage of a circuit.
- In other words, it can either step up (increase) or step down (decrease) the voltage.
- A transformer is necessary because sometimes the voltage requirements of different appliances are variable.
The current in the inductor __________________.
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 14 Detailed Solution
Download Solution PDFCONCEPT:
- Inductors: The coils of wire that are wound around any ferromagnetic material (iron cored) or wound around a hollow tube that increase their inductive value are called inductors.
- The inductance (L) is measured in Henry (H) and the instantaneous voltage in volts.
- Rate of instantaneous voltage is given by (v = L di/dt)
EXPLANATION:
- The given diagram is a simple inductor circuit with alternating current.
- From the phaser diagram, the Inductor current lags inductor voltage by 90° = π/2.
- The plot of current and voltage for this very simple circuit:
- From the current and voltage wave diagram, Inductor current lags inductor voltage by 90°. So option 1 is correct.
- Inductor current lags inductor voltage by 90°.
- Capacitor voltage lags current by 90°.
- In Resister only circuit, Voltage and Current are in the same phase.
- Or we can say there is no lag between current and voltage.
Value of current in an A.C. circuit is I = 2cos(ωt+θ). The value of Irms is:
Answer (Detailed Solution Below)
Electromagnetic Oscillations and Alternating Current Question 15 Detailed Solution
Download Solution PDFCONCEPT:
Alternating current:
- An alternating current can be defined as a current that changes its magnitude and polarity at a regular interval of time.
RMS value of current:
- RMS value of current is defined as the square root of the mean of the square of AC current for 1 complete cycle.
- It is equal to that value of DC current which generates the same heating effect as the AC current.
\(⇒ I_{rms}= \frac{I_{o}}{\sqrt{2}}\) -----(1)
Where Io = peak value of AC current
CALCULATION:
Given - I = 2cos(ωt+θ)
- The value of I will be maximum when cos(ωt+θ) is maximum,
- The maximum value of cos(ωt+θ) = 1
Therefore,
⇒ Io = 2A
So,
\(⇒ I_{rms}= \frac{I_{o}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}A\)
- Hence, option 1 is correct.