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For the circuit shown in fig. the input resistance is
Since op-amp is ideal
In the circuit of fig. the op-amp 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 Vio = 4 mV and Iio = 150 nA. The total output offset voltage is
io = ?
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 vi = -2 V, then output vo is
If vi < 0, then v0 > 0, D1 blocks and D2 conducts
Voltage follower v0 = v∞ = v+
The circuit shown in fig. is at steady state before the switch opens at t = 0. The vc(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 vi 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 op-amp is 60 dB. The magnitude of the vo is
The analog multiplier X of fig. has the characteristics vp = v1v2 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 vi > 2 V, output is positive. When vi < 2 V, output is negative.
In the op-amp circuit given in fig. the load current iL is
In the circuit of fig. output voltage is |vo| =1 V for a certain set of ω, R, an C. The |vo| 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 RF 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 Wien-bridge oscillator. The ratio is greater than 2. So there will be
In the circuit shown in fig. the op-amp is ideal. If βF = 60, then the total current supplied by the 15 V source is
v+ = 5V = v∞ = VE ,
The input current to the op-amp is zero.
In the circuit in fig. both transistor Q1 and Q2 are identical. The output voltage at T = 300 K is
In the op-amp series regulator circuit of fig. Vz = 62. V, VBE = 0.7 V and β = 60. The output voltage vo is