Introduction
In this article, we'll derive the expression for the charge on a capacitor in an RC circuit as a function of time. This is crucial for understanding transient behaviors in electrical circuits, which is important for JEE Mains and Advanced.
Theory
An RC circuit consists of a resistor (R) and a capacitor (C) connected in series with a voltage source. When the circuit is first connected, the capacitor starts charging through the resistor. The voltage across the capacitor increases over time until it reaches the source voltage.
Derivation
Consider the RC circuit with a voltage source \( V \), a resistor \( R \), and a capacitor \( C \). At any time \( t \), the voltage across the capacitor \( V_C(t) \) and the voltage across the resistor \( V_R(t) \) must add up to the source voltage:
\[ V = V_C(t) + V_R(t) \]
The voltage across the resistor is given by Ohm's law:
\[ V_R(t) = i(t) \cdot R \]
The current \( i(t) \) is the rate of change of charge \( Q(t) \) on the capacitor:
\[ i(t) = -\frac{dQ(t)}{dt} \]
Substitute \( i(t) \) into the resistor voltage equation:
\[ V_R(t) = -R \frac{dQ(t)}{dt} \]
The voltage across the capacitor is:
\[ V_C(t) = \frac{Q(t)}{C} \]
Thus, the equation for the voltage source becomes:
\[ V = \frac{Q(t)}{C} - R \frac{dQ(t)}{dt} \]
Rearranging gives:
\[ \frac{dQ(t)}{dt} + \frac{1}{RC}Q(t) = 0 \]
This is a first-order linear differential equation. Solving it yields:
\[ Q(t) = Q_0 \cdot e^{-\frac{t}{RC}} \]
where \( Q_0 \) is the initial charge on the capacitor at \( t = 0 \).
Conclusion
The derived formula \( Q(t) = Q_0 \cdot e^{-\frac{t}{RC}} \) shows that the charge on the capacitor decreases exponentially with time. Understanding this behavior helps in analyzing and designing circuits in practical applications.
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