The small box provided contains a cell of emf
E
and internal resistance
r
. Using this cell, an ammeter
(with negligible resistance), a variable (but known) resistance (
R
) and a plug key connect the circuit as
shown below.
Note
the meter polarity.
(a)
Does it matter in what order the components are connected? Explain.
(b)
In this experiment
I
is the dependent and
R
is the independent variable. Is this correct? Explain.
(c)
(d)
1
1
r
R
I
E
E
Write down the equation for
E
in terms of the current,
I
, the external resistance,
R
, and the
internal resistance of the cell,
r
.
(E2.1)
Show that equation (E2.1) can be rearranged into a form suitable for plotting a straight-line
graph:
As here question is referring a diagram but it is not attached. And question is also very splitted way, so it is not clear, hence i am taking the approximation.
As per the question,
Emf of cell = E
Internal resistance = r
Resistance of variable resistor = R
equivalent resistance of the circuit = R+r
current in the circuit "(I)=\\frac{E}{r+R}"
Potential developed across the internal resistance "E_r=rI = \\frac{rE}{R+r}"
Potential developed across the resistance R,
"V= IR = \\frac{ER}{r+R}"
And net emf across the battery = "V_{net}=E-\\frac{Er}{R+r}= \\frac{ER+Er-Er}{R+r}=\\frac{RE}{R+r}"
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