The correct answer is:
(a) It requires a large value of R and a small value of C
Explanation:
An integrator circuit is designed to produce an output that is proportional to the integral of the input signal with respect to time. In an ideal integrator, the output voltage Vout(t)V_{\text{out}}(t)Vout(t) is the integral of the input voltage Vin(t)V_{\text{in}}(t)Vin(t).
To understand why a low-pass RC circuit cannot be used as an integrator, let's look at the components:
- Low-pass RC circuit: This type of circuit typically consists of a resistor (R) and a capacitor (C). The cutoff frequency for an RC low-pass filter is given by fc=12πRCf_c = \frac{1}{2 \pi RC}fc=2πRC1.
- Integrator characteristics: An ideal integrator has a response where the output signal is the integral of the input. However, a low-pass filter primarily attenuates high-frequency signals and passes low-frequency signals, which is not the desired behavior for an integrator.
Why Option (a) is Correct:
- Large R and Small C: To make the integrator behavior work effectively, you need the time constant τ=RC\tau = RCτ=RC to be large. A large value of R and a small value of C will ensure that the capacitor charges and discharges over a longer time, allowing for the integration effect to occur.
- Small C allows the circuit to integrate at lower frequencies without causing a large phase shift or cutoff, while large R helps to slow down the response, promoting integration.
Why the Other Options Are Incorrect:
- (b) Large value of C and small value of R: This would make the time constant too small, causing the circuit to behave more like a high-pass filter instead of an integrator.
- (c) Large value of R and C: This could cause an excessive time constant, which might affect the stability of the integrator.
- (d) Small value of R and C: This would result in a very fast response, not suited for integration.
Thus, to create an integrator, an RC circuit requires a large value of R and a small value of C, making option (a) the correct choice.