Question:

Please convert the below content to rendered equations without any errors.Alright, let’s go **step by step** and keep everything clean and logical. --- ## 1) Understanding the circuit * The **op-amp is ideal** * Infinite gain * Infinite input impedance * Zero output impedance * The **non-inverting (+) input is grounded** * Therefore, due to negative feedback: [ V_- = V_+ = 0 \quad \text{(virtual ground)} ] So the node connecting **L, R, and the inverting input** is at **0 V**. --- ## 2) Label the key node Let the node between the inductor (L), resistor (R), and the inverting input be node **A**. [ V_A = 0 ] --- ## 3) Apply KCL at node A Because the op-amp input draws **no current**, the current through the inductor must equal the current through the resistor. [ i_L = i_R ] --- ## 4) Current through the inductor Voltage across the inductor: [ V_L = V_{in}(t) - V_A = V_{in}(t) ] Inductor current relation: [ V_L = L \frac{di_L}{dt} ] So, [ i_L = \frac{1}{L} \int V_{in}(t), dt ] --- ## 5) Current through the resistor Voltage across the resistor: [ V_R = V_A - V_{out}(t) = -V_{out}(t) ] Resistor current: [ i_R = \frac{V_R}{R} = -\frac{V_{out}(t)}{R} ] --- ## 6) Equate the currents [ \frac{1}{L} \int V_{in}(t), dt = -\frac{V_{out}(t)}{R} ] --- ## 7) Solve for (V_{out}(t)) [ V_{out}(t) = -\frac{R}{L} \int V_{in}(t), dt ] --- ## 8) Final expression (time domain) [ \boxed{V_{out}(t) = -\frac{R}{L} \int V_{in}(t), dt} ] --- ## 9) Interpretation * The circuit behaves as an **ideal integrator** * The **negative sign** comes from the inverting configuration * The gain factor is (\frac{R}{L}) --- ## 10) (Optional) Laplace-domain form Taking Laplace transforms: [ \frac{V_{out}(s)}{V_{in}(s)} = -\frac{R}{sL} ] --- ### ✅ Final takeaway This circuit is an **op-amp based integrator using an inductor at the input**, where the output voltage is proportional to the **integral of the input voltage** scaled by (R/L). If you want, I can also: * Verify with **frequency response** * Explain **physical intuition** * Compare it with **RC integrator** Just tell me ?