Question:
In an inverting RC integrator using an Op-amp, \( RC = 1\,\text{s}, \, V_i = 5\,\text{V} \), then \( V_o = \, \_\_\_\_\_\_\_ \)
In an inverting RC integrator using an Op-amp, \( RC = 1\,\text{s}, \, V_i = 5\,\text{V} \), then \( V_o = \, \_\_\_\_\_\_\_ \)
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For constant input, integrator output is a ramp: \( V_o = -V_i t / RC \)
Updated On: Jun 24, 2025
- \( 10t \)
- \( -5t \)
- \( 5t \)
- \( 15t \)
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The Correct Option is B
Solution and Explanation
Step 1: Output of inverting integrator
The voltage output of an inverting op-amp integrator is: \[ V_o(t) = -\frac{1}{RC} \int V_i(t) dt \] Step 2: Substitute given values
Given:
- \( RC = 1 \)
- \( V_i(t) = 5 \, \text{V (constant)} \)
So: \[ V_o(t) = -\frac{1}{1} \int 5 dt = -5t \] Hence, the correct output is Option (2).
The voltage output of an inverting op-amp integrator is: \[ V_o(t) = -\frac{1}{RC} \int V_i(t) dt \] Step 2: Substitute given values
Given:
- \( RC = 1 \)
- \( V_i(t) = 5 \, \text{V (constant)} \)
So: \[ V_o(t) = -\frac{1}{1} \int 5 dt = -5t \] Hence, the correct output is Option (2).
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