Final answer:
When a capacitor is connected to a battery and resistor, it begins charging, with its voltage initially rising rapidly then approaching the battery's emf asymptotically. The voltage across the capacitor eventually becomes equal to the emf when reaching a steady state, and the voltage across the resistor drops to zero as no current flows.
Step-by-step explanation:
When we connect a battery and a resistor across a capacitor, the voltage across the capacitor changes over time due to the charging process.
Initially, the voltage across the capacitor (Vc) is zero, and it begins to rise rapidly because the initial current is at its maximum.
As time progresses, the voltage increases and approaches the electromotive force (emf) of the battery asymptotically according to the formula V = emf(1 - e-t/RC), where t is the time, R is the resistance, and C is the capacitance.
The rise in the voltage across the capacitor is due to the accumulation of charge on the plates of the capacitor. As the charge increases, the current through the circuit decreases along with the voltage difference across the resistor, which can be represented by the formula
VR(t) = (IR) e-t/τ = εe-t/τ, where ε is the battery voltage and τ (tau) is the time constant of the RC circuit.
On reaching a steady state, the voltage across the capacitor is equal to the emf of the battery, and the voltage across the resistor becomes zero because no current flows through the resistor at that point.