the internal resistance of a cell of emf 2V is 0.1ohm it is connected to a resistance of 0.9ohm. the voltage across the cell will be....... A)0.5v B)1.8V C)1.95V D)3V
Answers
Answer:
1.95V
Explanation:
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.
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Answer:
A resistor of resistance 3.9Ω is connected in series with internal resistance of cell 0.1Ω.
Total resistance of the circuit R
′
=0.1+3.9=4.0Ω
Emf of cell E=2V
Current flowing in the circuit I=
R
′
E
∴ I=
4
2
=0.5A
Thus voltage across the cell V
cell
=IR where R=3.9Ω
⟹ V
cell
=0.5×3.9=1.95V
Explanation