Question

A cell of emf 2 volt and internal resistance 0.1 ohm is connected to a 3.9 ohm external resistance what will be the potential difference across the terminal of the cell

Answer 1

Final Answer : 1.95V

Underst\nding :
1) Potential Difference across terminal of a cell,
V = E - I r
where
E = EMF of cell =2V
r = internal resistance =0.1ohm
i = current in circuit.
R = External Resistor

Steps :
1) Cell and External resistor are connected directly,
=> R(eq) = r + R
=0.1 + 3.9 = 4.0 oh m

2) Current in circuit,
I = E / R(eq)
= 2/4.0 = 1/2 =0.5 A


3) Potential Difference across terminal of a cell,
V = E - Ir
= 2 - 0.5 *0.1
= 1.95 V.


See Circuit Diagram in pic

Answer 2

Answer:

Given,

EMF of the cell(E) =2 V

internal resistance(r)= 0.1 V

External resistance(R)=3.9 V

V=IR

=(E÷(R+r))×R

=(2÷(3.9+0.1))×3.9

=1.95 V

Therefore potential difference across the terminals of the cell =

1.95V.

Explanation: