In a pure ……………….. Circuit current leads the voltage by `pi` /2 .

Answer

To solve the question, "In a pure ……………….. Circuit current leads the voltage by `pi` /2," we need to identify the type of circuit being referred to. The question hints at a specific relationship between current and voltage in an AC circuit.

### Step-by-Step Solution:

1. **Understanding the Phase Relationship**: In AC circuits, the phase relationship between current and voltage is crucial. The question states that the current leads the voltage by `pi/2`. This means that the current reaches its maximum value a quarter of a cycle (or 90 degrees) before the voltage does.

2. **Identifying the Circuit Type**: The phase relationship where the current leads the voltage by `pi/2` is characteristic of a capacitive circuit. In a pure capacitive circuit, the current leads the voltage due to the nature of capacitors storing and releasing energy.

3. **Mathematical Representation**: In a capacitive circuit, if we denote the voltage as \( V = V_0 \sin(\omega t) \), the current can be expressed as:
\[ I = C \frac{dV}{dt} \]
where \( C \) is the capacitance. When we differentiate the voltage with respect to time, we find that the current is:
\[ I = C \cdot \omega V_0 \cos(\omega t) = C \cdot \omega V_0 \sin\left(\omega t + \frac{\pi}{2}\right) \]
This shows that the current is leading the voltage by `pi/2`.

4. **Conclusion**: Thus, we conclude that in a pure capacitive circuit, the current leads the voltage by `pi/2`.

### Final Answer:
In a pure **capacitive** circuit, current leads the voltage by `pi/2`.

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