A 20 nF capacitor is connected across an AC generator that produces a peak voltage of 5.0 V. What is the instantaneous value of the emf at the instant when i_C = I_C?

Verified step by step guidance

1. Step 1: Understand the problem. The question involves an AC circuit with a capacitor, and it asks for the instantaneous value of the emf (electromotive force) when the current through the capacitor equals the peak current. This requires understanding the relationship between voltage, current, and phase in an AC circuit with a capacitor.
2. Step 2: Recall the relationship between current and voltage in a capacitor. In an AC circuit, the current leads the voltage by 90 degrees (π/2 radians) in phase. The peak current (I_C) is related to the peak voltage (V_peak) by the capacitive reactance (X_C). The formula for capacitive reactance is: $X_C=\frac{1}{\omega C_{}}$, where ω is the angular frequency and C is the capacitance.
3. Step 3: Write the expression for the instantaneous emf in an AC circuit. The emf varies sinusoidally with time and is given by: $V=V_{\mathrm{peak}}\cdot \sin \left(\omega t\right)$. Here, V_peak is the peak voltage, ω is the angular frequency, and t is the time.
4. Step 4: Relate the current to the emf. The current through the capacitor is given by: $i_C=I_{\mathrm{peak}}\cdot \sin (\omega t+\frac{\pi }{2})$. At the instant when i_C = I_C (peak current), the emf will correspond to the phase difference between voltage and current.
5. Step 5: Solve for the instantaneous emf. Substitute the time t corresponding to the peak current into the voltage equation. Since the current leads the voltage by 90 degrees, the phase difference must be accounted for. Use the sinusoidal relationship to find the instantaneous emf value at this specific time.

Key Concepts

Capacitance

Capacitance is the ability of a capacitor to store electrical charge per unit voltage. It is measured in farads (F), and in this case, the capacitor has a capacitance of 20 nF (nanofarads). The relationship between charge (Q), voltage (V), and capacitance (C) is given by the formula Q = C * V, which is essential for understanding how capacitors behave in circuits.

AC Voltage and Peak Voltage

AC (alternating current) voltage varies sinusoidally over time, characterized by its peak voltage, which is the maximum voltage reached in one cycle. In this scenario, the peak voltage of the AC generator is 5.0 V. Understanding the instantaneous voltage at any point in time requires knowledge of the sine function, as the voltage can be expressed as V(t) = V_peak * sin(ωt), where ω is the angular frequency.

Instantaneous Current in a Capacitor

The instantaneous current (i_C) through a capacitor in an AC circuit is related to the rate of change of voltage across it. It can be expressed as i_C = C * (dV/dt), where dV/dt is the derivative of the voltage with respect to time. At the moment when i_C equals the peak current, understanding the phase relationship between current and voltage in capacitors is crucial, as the current leads the voltage by 90 degrees in a purely capacitive circuit.