Question

A capacitor is connected to a cell of emf E and some internal resistance the potential difference the capacitor it is fully charged is –
A. Cell is E
B. Cell is $< E$
C. Capacitor is $< E$
D. Capacitor is $> E$

Answer 1

We know that when the capacitor is fully charged, it draws no current.
We can use ohm's law.$V = IR$

Complete answer:
A capacitor is a device that stores electrical energy in an electric field; it is a passive electronic component with two terminals.Potential difference is the difference of electrical potential bather two points. The potential difference of a supply is a measure of the energy given to the charge carried in a circuit. Unlike current which pious around a closed electrical circuit in the form of electrical charge,

When the capacitor is fully charged,it draws no current. When the capacitor is fully charged, it draws no current. When no current flows in the circuit, potential difference across the cell =potential difference across the capacitor.In this problem this is open circuit so voltage acres the cell is equal to itself.In the given problem says that the potential difference p.d have to find when the capacitor is fully charged.

$\therefore$Potential difference is equal to E.

So, the correct answer is “Option A”.

Note:
We need to know when capacitor is fully charged no current flows throw so the electrical current$i = 0$ as voltage is Potential difference is proportional to I
$V\alpha I \\\ V = IR \\\ V = 0.R \\\ V = 0 \\\$