A cell has an e.m.f. of 1.08 V and an internal resistance of 0.5 ohm .When it is connected in series with resistor of R, the potential difference between the terminal fell to 0.96 V .What was the value of R ?
Answers
The current through the internal resistance drops the voltage by
(1.08 - 0.96) volts = 0.12 volts
I = Edropped/Rint = 0.12/0.5 = 0.24 amps
The load resistance has 0.96v across it when passing 0.24 amps.
Rext = I/E = 0.96/0.24
= 4 ohms
Answer:
it is given that the potential drop across the cell is
0.96
V
; that means this drop is due to some potential drop across the
0.5
Ω
internal resistance in the cell, we can say the rest i.e
(
1.08
−
0.96
)
=
0.12
V
has dropped across
0.5
Ω
.
So, if the current flowing through the circuit is
I
, then we can say,
I
×
0.5
=
0.12
Or,
I
=
0.24
A
Now, the voltage drop across resistor
R
is
(
1.08
−
0.12
)
=
0.96
V
(note it is the same as
V
A
−
V
B
)
So,if voltage drop across
R
is
0.96
V
and Current flowing is
0.24
A
Then, using Ohm's Law, we can write,
R
=
0.96
0.24
=
4
Ω: