"MCQ Test on Electromagnetic Induction Fundamentals"
Time Left: 45:00
Total Marks: 60, Obtained Marks: 0
1. In an LC circuit, the capacitor has maximum charge \( q_0 \). The value of \( \left( \frac{dI}{dt} \right)_{\text{max}} \) is:
A. \( \frac{q_0}{LC} \)
B. \( \frac{q_0}{\sqrt{LC}} \)
C. \( \frac{q_0}{\sqrt{LC}} - 1 \)
D. \( \frac{q_0}{LC} + 1 \)
The correct answer is \( \frac{q_0}{LC} \). Here's how we derive it:
Step 1: Basic LC Circuit Relationship
In an LC circuit, the charge on the capacitor as a function of time is given by:
\( q(t) = q_0 \cos(\omega t) \), where \( \omega = \frac{1}{\sqrt{LC}} \).
Step 2: Current in the Circuit
The current is the time derivative of the charge:
\( I(t) = \frac{dq(t)}{dt} = -q_0 \omega \sin(\omega t) \). The maximum current is:
\( I_{\text{max}} = \frac{q_0}{\sqrt{LC}} \).
Step 3: Rate of Change of Current
Now, we find the time derivative of the current:
\( \frac{dI(t)}{dt} = -q_0 \omega^2 \cos(\omega t) \).
The maximum value of \( \frac{dI}{dt} \) is \( \frac{q_0}{LC} \), which occurs when \( \cos(\omega t) = 1 \).
Step 1: Basic LC Circuit Relationship
In an LC circuit, the charge on the capacitor as a function of time is given by:
\( q(t) = q_0 \cos(\omega t) \), where \( \omega = \frac{1}{\sqrt{LC}} \).
Step 2: Current in the Circuit
The current is the time derivative of the charge:
\( I(t) = \frac{dq(t)}{dt} = -q_0 \omega \sin(\omega t) \). The maximum current is:
\( I_{\text{max}} = \frac{q_0}{\sqrt{LC}} \).
Step 3: Rate of Change of Current
Now, we find the time derivative of the current:
\( \frac{dI(t)}{dt} = -q_0 \omega^2 \cos(\omega t) \).
The maximum value of \( \frac{dI}{dt} \) is \( \frac{q_0}{LC} \), which occurs when \( \cos(\omega t) = 1 \).
2. The natural frequency of the circuit shown in figure is:
A. \( \frac{1}{\sqrt{LC}} \)
B. \( \frac{1}{\sqrt{2LC}} \)
C. \( \frac{2}{\sqrt{LC}} \)
D. None of these
The correct answer is \( \frac{1}{\sqrt{LC}} \). The two inductors \( L \) are in series, giving an equivalent inductance of \( 2L \). Similarly, the two capacitors \( C \) are in series, giving an equivalent capacitance of \( \frac{C}{2} \).
Using the formula for the natural frequency, \( \omega_0 = \frac{1}{\sqrt{L_{\text{eq}} C_{\text{eq}}}} \), and substituting the equivalent values \( L_{\text{eq}} = 2L \) and \( C_{\text{eq}} = \frac{C}{2} \), we find that:
\( \omega_0 = \frac{1}{\sqrt{LC}} \).
Using the formula for the natural frequency, \( \omega_0 = \frac{1}{\sqrt{L_{\text{eq}} C_{\text{eq}}}} \), and substituting the equivalent values \( L_{\text{eq}} = 2L \) and \( C_{\text{eq}} = \frac{C}{2} \), we find that:
\( \omega_0 = \frac{1}{\sqrt{LC}} \).
3. In an AC generator, a coil with \( N \) turns, all of the same area \( A \) and total resistance \( R \), rotates with angular frequency \( \omega \) in a magnetic field \( B \). The maximum value of emf generated in the coil is:
A. \( NAB \)
B. \( NABR \)
C. \( NAB\omega \)
D. None of these
The correct answer is \( NAB\omega \). According to Faraday's law of electromagnetic induction, the maximum EMF generated in a rotating coil is given by \( \mathcal{E}_{\text{max}} = NAB\omega \), where \( N \) is the number of turns, \( A \) is the area, \( B \) is the magnetic field strength, and \( \omega \) is the angular velocity of the coil.
4. A transformer having efficiency of 90% is working on 200V and 3kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
A. 450 V, 15 A
B. 300 V, 12 A
C. 500 V, 10 A
D. 600 V, 20 A
The correct answers are 450 V for the voltage across the secondary coil and 15 A for the current in the primary coil. The output power of the transformer is calculated using its efficiency, leading to a voltage across the secondary coil found through the formula \( V_s = \frac{P_{\text{output}}}{I_s} \). The current in the primary coil is determined by \( I_p = \frac{P_{\text{input}}}{V_p} \).
5. An L-C circuit contains a 0.60H inductor and a 25 µF capacitor. What is the rate of change of the current (in amp/second) when the charge on the capacitor is \(3.0 \times 10^{-5}\) C?
A. 2.0
B. 4.0
C. 3.0
D. 6.0
The correct answer is 2.0 A/s.
Given data:
Inductance \( L = 0.60 \, \text{H} \), Capacitance \( C = 25 \, \mu \text{F} = 25 \times 10^{-6} \, \text{F} \), Charge \( Q = 3.0 \times 10^{-5} \, \text{C} \).
Using the relation \( L \frac{dI}{dt} = - \frac{Q}{C} \), we calculate the rate of change of current:
\[ \frac{dI}{dt} = - \frac{Q}{LC} \] Substituting the values: \[ \frac{dI}{dt} = - \frac{3.0 \times 10^{-5}}{(0.60)(25 \times 10^{-6})} \] Simplifying: \[ \frac{dI}{dt} = - \frac{3.0 \times 10^{-5}}{1.5 \times 10^{-5}} = -2 \, \text{A/s} \] Thus, the rate of change of current is 2.0 A/s.
Given data:
Inductance \( L = 0.60 \, \text{H} \), Capacitance \( C = 25 \, \mu \text{F} = 25 \times 10^{-6} \, \text{F} \), Charge \( Q = 3.0 \times 10^{-5} \, \text{C} \).
Using the relation \( L \frac{dI}{dt} = - \frac{Q}{C} \), we calculate the rate of change of current:
\[ \frac{dI}{dt} = - \frac{Q}{LC} \] Substituting the values: \[ \frac{dI}{dt} = - \frac{3.0 \times 10^{-5}}{(0.60)(25 \times 10^{-6})} \] Simplifying: \[ \frac{dI}{dt} = - \frac{3.0 \times 10^{-5}}{1.5 \times 10^{-5}} = -2 \, \text{A/s} \] Thus, the rate of change of current is 2.0 A/s.
6. A capacitor of capacity \( 2 \, \mu F \) is charged to a potential difference of 12V. It is then connected across an inductor of inductance 0.6 mH. What is the current in the circuit at a time when the potential difference across the capacitor is 6V?
A. 0.3 A
B. 0.6 A
C. 0.9 A
D. 1.2 A
The correct answer is 0.6 A.
The total energy in the LC circuit is conserved, and is initially stored in the capacitor. We use the energy conservation formula:
\[ U = \frac{1}{2} C V_0^2 = \frac{1}{2} L I^2 + \frac{1}{2} C V^2 \] where \( V_0 = 12V \), \( V = 6V \), \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \), and \( L = 0.6 \, mH = 0.6 \times 10^{-3} \, H \).
Step 1: Calculate the initial energy stored in the capacitor:
\[ U = \frac{1}{2} C V_0^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 12^2 = 1.44 \times 10^{-4} \, J \] Step 2: Write the energy conservation equation: \[ 1.44 \times 10^{-4} = \frac{1}{2} \times 0.6 \times 10^{-3} \times I^2 + \frac{1}{2} \times 2 \times 10^{-6} \times 6^2 \] Step 3: Simplify and solve for \( I \): \[ 1.08 \times 10^{-4} = 0.3 \times 10^{-3} \times I^2 \quad \Rightarrow \quad I = \sqrt{0.36} = 0.6 \, A \] Thus, the current in the circuit is 0.6 A.
The total energy in the LC circuit is conserved, and is initially stored in the capacitor. We use the energy conservation formula:
\[ U = \frac{1}{2} C V_0^2 = \frac{1}{2} L I^2 + \frac{1}{2} C V^2 \] where \( V_0 = 12V \), \( V = 6V \), \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \), and \( L = 0.6 \, mH = 0.6 \times 10^{-3} \, H \).
Step 1: Calculate the initial energy stored in the capacitor:
\[ U = \frac{1}{2} C V_0^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 12^2 = 1.44 \times 10^{-4} \, J \] Step 2: Write the energy conservation equation: \[ 1.44 \times 10^{-4} = \frac{1}{2} \times 0.6 \times 10^{-3} \times I^2 + \frac{1}{2} \times 2 \times 10^{-6} \times 6^2 \] Step 3: Simplify and solve for \( I \): \[ 1.08 \times 10^{-4} = 0.3 \times 10^{-3} \times I^2 \quad \Rightarrow \quad I = \sqrt{0.36} = 0.6 \, A \] Thus, the current in the circuit is 0.6 A.
7. In the circuit shown, the switch has been in position 1 for a long time. Now the switch is shifted to position 2. If this instant is taken as \( t = 0 \), then at time \( t = \frac{\pi}{2} \), what is the current in the circuit?
A. 0 A
B. 2 A
C. 4 A
D. 6 A
The correct answer is 4 A.
Initially, the inductor behaves like a short circuit after the switch remains in position 1 for a long time. The initial current through the inductor is given by Ohm's law:
\[ I_0 = \frac{V}{R} = \frac{20}{5} = 4 \, \text{A} \] When the switch is moved to position 2, the circuit forms an L-C-R circuit. At \( t = \frac{\pi}{2} \), the current can be found based on the initial conditions of the circuit. Given the natural frequency \( \omega_0 = 1 \, \text{rad/s} \) and the damping factor \( \zeta = 2.5 \), the system is overdamped. However, the initial current remains 4 A at \( t = 0 \), and it decreases over time.
Initially, the inductor behaves like a short circuit after the switch remains in position 1 for a long time. The initial current through the inductor is given by Ohm's law:
\[ I_0 = \frac{V}{R} = \frac{20}{5} = 4 \, \text{A} \] When the switch is moved to position 2, the circuit forms an L-C-R circuit. At \( t = \frac{\pi}{2} \), the current can be found based on the initial conditions of the circuit. Given the natural frequency \( \omega_0 = 1 \, \text{rad/s} \) and the damping factor \( \zeta = 2.5 \), the system is overdamped. However, the initial current remains 4 A at \( t = 0 \), and it decreases over time.
8. A coil of \( N \) turns and area \( A \) is placed in a uniform magnetic field \( B \). If the magnetic field is decreased to zero in \( t \) seconds, the induced EMF in the coil is:
A. \( \frac{N \cdot A \cdot B}{t} \)
B. \( \frac{N \cdot B \cdot A}{t} \)
C. \( \frac{N \cdot A \cdot B}{t^2} \)
D. \( \frac{N \cdot A \cdot B^2}{t} \)
The correct answer is \( \frac{N \cdot A \cdot B}{t} \). This follows from Faraday's law of electromagnetic induction, where the induced EMF is proportional to the rate of change of the magnetic flux through the coil.
9. A circular loop of radius \( r \) is placed in a uniform magnetic field \( B \) perpendicular to the plane of the loop. If the magnetic field is increased at a rate of \( \frac{dB}{dt} \), the induced EMF in the loop is:
A. \( \pi r^2 \frac{dB}{dt} \)
B. \( \frac{1}{2} \pi r^2 \frac{dB}{dt} \)
C. \( 2\pi r^2 \frac{dB}{dt} \)
D. \( \frac{dB}{dt} \)
The correct answer is \( \pi r^2 \frac{dB}{dt} \). The induced EMF is calculated using Faraday's law as the rate of change of magnetic flux. The magnetic flux is proportional to the area of the loop, which gives the factor \( \pi r^2 \).
10. A solenoid with \( N \) turns and length \( L \) carries a current \( I \). If the current is increased at a rate of \( \frac{dI}{dt} \), the induced EMF in the solenoid is:
A. \( -\frac{N \mu_0 A}{L} \frac{dI}{dt} \)
B. \( \frac{N \mu_0 A}{L} \frac{dI}{dt} \)
C. \( \frac{N \mu_0 L}{A} \frac{dI}{dt} \)
D. \( -\frac{L}{N \mu_0 A} \frac{dI}{dt} \)
The correct answer is \( \frac{N \mu_0 A}{L} \frac{dI}{dt} \). The EMF is induced due to the change in current, which affects the magnetic flux. This expression is derived from the solenoid's self-inductance formula.
11. A rectangular coil of area \( A \) with \( N \) turns rotates in a magnetic field \( B \) at an angular frequency \( \omega \). The maximum induced EMF in the coil is:
A. \( NAB\omega \)
B. \( \frac{NAB\omega}{2} \)
C. \( N \cdot \frac{BA^2}{\omega} \)
D. \( NAB \)
The correct answer is \( NAB\omega \). This formula is derived from Faraday's law for rotating coils, where the rate of change of flux generates the EMF.
12. In an AC generator, a coil with \( N \) turns, all of the same area \( A \) and total resistance \( R \), rotates with \( \omega \) frequency in a magnetic field \( B \). The maximum value of emf generated in the coil is:
A. \( NABR \)
B. \( NAB \)
C. \( \frac{NAB}{R} \)
D. \( \frac{NABR}{R} \)
The correct answer is \( NAB \). The maximum induced EMF in a coil rotating in a magnetic field is given by Faraday's law of induction, where the induced EMF is directly proportional to the magnetic flux change, represented here as \( NAB \).
13. A transformer having an efficiency of 90% is working on 200V and 3kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
A. 100 V and 12 A
B. 150 V and 8 A
C. 120 V and 10 A
D. 200 V and 6 A
The correct answer is 100 V and 12 A. The power in the secondary coil is \( P_s = V_s \times I_s \). With an efficiency of 90%, we can calculate the primary voltage and current based on the given values.
14. The natural frequency of the circuit shown in figure is:
A. \( \frac{1}{\sqrt{LC}} \)
B. \( \frac{1}{\sqrt{2LC}} \)
C. \( \frac{2}{\sqrt{LC}} \)
D. None of these
The correct answer is \( \frac{1}{\sqrt{LC}} \). The natural frequency of an LC circuit is determined by the inductance \( L \) and capacitance \( C \), where the formula represents the resonant frequency.
15. In an LC circuit, the capacitor has maximum charge \( q_0 \). The value of \( \left( \frac{dI}{dt} \right)_{max} \) is:
A. \( \frac{q_0}{LC} \)
B. \( \frac{q_0}{\sqrt{LC}} \)
C. \( \frac{q_0}{LC} - 1 \)
D. \( \frac{q_0}{LC} + 1 \)
The correct answer is \( \frac{q_0}{LC} \). The rate of change of current at maximum charge is related to the maximum charge and the inductance-capacitance product.
.png)
0 Comments