Question:
The current gain \( \left( \frac{I_{out}}{I_{in}} \right) \) in the circuit with an ideal current amplifier given below is
The current gain \( \left( \frac{I_{out}}{I_{in}} \right) \) in the circuit with an ideal current amplifier given below is
Show Hint
In a current amplifier circuit, the current gain is given by the ratio of the input capacitance to the feedback capacitance.
Updated On: Nov 27, 2025
- ( \frac{C_f}{C_c} \)
- ( \frac{-C_f}{C_c} \)
- ( \frac{C_c}{C_f} \)
- ( \frac{-C_c}{C_f} \)
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the current amplifier circuit.
In the given circuit with an ideal current amplifier, the feedback capacitance \( C_f \) and the input capacitance \( C_c \) determine the current gain. The current gain of a current amplifier is the ratio of the output current to the input current, which depends on the ratio of these capacitances. Step 2: Analyzing the options.
- (A) \( \frac{C_f}{C_c} \): This is not the correct expression for the current gain.
- (B) \( \frac{-C_f}{C_c} \): This is also incorrect since the negative sign is not applicable in this case.
- (C) \( \frac{C_c}{C_f} \): This is the correct expression for the current gain in the given ideal current amplifier circuit.
- (D) \( \frac{-C_c}{C_f} \): This is incorrect for similar reasons as (B).
Step 3: Conclusion.
The correct answer is (C), \( \frac{C_c}{C_f} \), which represents the current gain of the circuit.
In the given circuit with an ideal current amplifier, the feedback capacitance \( C_f \) and the input capacitance \( C_c \) determine the current gain. The current gain of a current amplifier is the ratio of the output current to the input current, which depends on the ratio of these capacitances. Step 2: Analyzing the options.
- (A) \( \frac{C_f}{C_c} \): This is not the correct expression for the current gain.
- (B) \( \frac{-C_f}{C_c} \): This is also incorrect since the negative sign is not applicable in this case.
- (C) \( \frac{C_c}{C_f} \): This is the correct expression for the current gain in the given ideal current amplifier circuit.
- (D) \( \frac{-C_c}{C_f} \): This is incorrect for similar reasons as (B).
Step 3: Conclusion.
The correct answer is (C), \( \frac{C_c}{C_f} \), which represents the current gain of the circuit.
Was this answer helpful?
0
0
Top Questions on Amplifiers
- A continuous-time periodic signal \( x(t) = 1 + 2 \cos(4\pi t) + 2 \cos(6\pi t) \). If \( T \) is the time period of \( x(t) \), then \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt = \ \_\_\_\_\_? \]
- The output impedance of a non-ideal operational amplifier is denoted by \( Z_{out} \). The variation in the magnitude of \( Z_{out} \) with increasing frequency, \( f \), in the circuit shown below, is best represented by:
- For an ideal MOSFET biased in saturation, the magnitude of the small signal current gain for a common drain amplifier is
Questions Asked in GATE EE exam
- A 3-phase, 400 V, 4 pole, 50 Hz star-connected induction motor has the following parameters referred to stator: \[ R'_1 = 1 \, \text{ohm}, \, X_s = X'_1 = 2 \, \text{ohm} \] Stator resistance, magnetizing reactance, and core loss of the motor are neglected. The motor is run with constant \( v/f \) control. The output frequency from the drive is closest to:
- GATE EE - 2025
- Transient and Steady‐state analysis of linear time invariant systems
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in V/Ω, is __________ (round off to two decimal places).
- GATE EE - 2025
- Electrical Energy, Power
- A DC series motor with negligible series resistance is running at a certain speed driving a load, where the load torque varies as the cube of the speed. The motor is fed from a 400 V DC source and draws 40 A armature current. Assume a linear magnetic circuit. The external resistance in ohms that must be connected in series with the armature to reduce the speed of the motor by half is closest to:
- GATE EE - 2025
- Instrument transformers
- The input voltage \( v(t) \) and current \( i(t) \) of a converter are given by, \[ v(t) = 300 \sin(\omega t) \, {V} \] \[ i(t) = 10 \sin\left(\omega t - \frac{\pi}{6}\right) + 2 \sin\left(3\omega t + \frac{\pi}{6}\right) + \sin\left(5\omega t + \frac{\pi}{2}\right) \, {A} \] where \( \omega = 2\pi \times 50 \) rad/s. The input power factor of the converter is closest to:
- Instrument(s) required to synchronize an alternator to the grid is/are:
- GATE EE - 2025
- Electrical Energy, Power
View More Questions