For the circuit shown in fig. the input resistance is
Since opamp is ideal
In the circuit of fig. the opamp slew rate is SR = 0.5 V/μs. If the amplitude of input signal is 0.02 V, then the maximum frequency that may be used is
In the circuit of fig. the input offset voltage and input offset current are V_{io} = 4 mV and I_{io} = 150 nA. The total output offset voltage is
i_{o} = ?
If ‘V’ is the voltage phasor and ‘I’ is the current phasor, then VI represents
Apparent Power (S): It is defined as the product of r.m.s value of voltage (V) and current (1). It is denoted by S. S = V/I Volt Ampere
Consider the circuit shown below
Que: If v_{i} = 2 V, then output v_{o} is
If v_{i} < 0, then v_{0} > 0, D_{1} blocks and D_{2} conducts
Voltage follower v_{0} = v_{∞} = v_{+}
The circuit shown in fig. is at steady state before the switch opens at t = 0. The v_{c}(t) for t > 0 is
For t > 0 the equivalent circuit is shown in fig.
The LED in the circuit of fig. will be on if v_{i }is
When v_{+} > 5 V, output will be positive and LED will be on. Hence (C) is correct.
In the circuit of fig. the CMRR of the opamp is 60 dB. The magnitude of the v_{o} is
The analog multiplier X of fig. has the characteristics v_{p} = v_{1}v_{2} The output of this circuit is
v_{+} = 0 = v_{}
If the input to the ideal comparator shown in fig. is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of
When v_{i} > 2 V, output is positive. When v_{i} < 2 V, output is negative.
In the opamp circuit given in fig. the load current i_{L} is
In the circuit of fig. output voltage is v_{o} =1 V for a certain set of ω, R, an C. The v_{o} will be 2 V if
This is a all pass circuit
Thus when ω and R is changed, the transfer function is unchanged
In the circuit of fig. the 3 dB cutoff frequency is
The phase shift oscillator of fig. operate at f = 80 kHz. The value of resistance R_{F} is
The oscillation frequency is
The value of C required for sinusoidal oscillation of frequency 1 kHz in the circuit of fig. is
This is Wienbridge oscillator. The ratio is greater than 2. So there will be
oscillation
In the circuit shown in fig. the opamp is ideal. If β_{F} = 60, then the total current supplied by the 15 V source is
v_{+} = 5V = v_{∞ }= V_{E} ,
The input current to the opamp is zero.
In the circuit in fig. both transistor Q_{1} and Q_{2} are identical. The output voltage at T = 300 K is
In the opamp series regulator circuit of fig. V_{z} = 62. V, V_{BE} = 0.7 V and β = 60. The output voltage v_{o} is
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 








