To find the charge on the capacitor in the given circuit in steady state, we need to understand some important concepts and steps. In steady state, the capacitor acts as an open circuit for DC currents.
First, identify that in a steady state of a DC circuit, the capacitor becomes fully charged and no current flows through it, which makes it act like an open circuit. Hence, the voltage across the capacitor is the same as the voltage across the open circuit.
Let's consider that the diode in forward bias allows the current through the circuit, having an effect on the resistors and the capacitor charging process.
The voltage across the capacitor in steady state will be equal to the total voltage applied from the source minus any voltage drops across other components directly in series with it.
In this circuit:
The capacitor's voltage can be directly found from the voltage across the parallel combination of resistors it is connected to, assuming ideal diodes and steady state condition, meaning the drop across 1 Ω will be zero.
The formula for the charge \( Q \) on a capacitor is given by:
\(Q = C \times V\)
Where \( C \) is the capacitance of the capacitor, \( 5 \, \mu \text{F} \).
Assume that in steady state conditions, the voltage drop across the capacitor equals the full applied voltage, \( 2.5 \, \text{V} \).
