Page 57 - Physics_Form_2

Current electricity

Suppose the switch is closed as plotted as shown in Figure 20.20 based on the
shown in Figure 2.19 (b), current, equation:
I, flows through the circuit. From 
1
Ohm’s law, the p.d, V, across the R  E   r
resistor, R, is given by, I 
FOR ONLINE READING ONLY
With R in the vertical axis and the reciprocal of
V = IR
current in the horizontal axis.
The potential drop (V ) caused
1
by the cell’s internal resistance is R (Ω)
given by,

V = Ir
1
r
The total voltage in the circuit is 1 (A )
−1
I
the sum of the potential drop due E
to the cell’s internal resistance
and that of the resistor, R. That is,
-r (Ω)
E V= + Ir

When no current is drawn in
the circuit, e.m.f is equal to p.d Figure 2.20: The graph of resistance, R,
across the cell EV= . Generally, against the reciprocal of current, 1
the total e.m.f of the cell (s) is I
given by, The numerical value of e.m.f and r can be

E V V= + 1 determined from the graph. The slope of the

= IR + Ir graph represents the value of e.m.f, E, and the
vertical intercept represents the value of internal
E = ( IR r+ ) resistance, r, of the cell. Similarly, when the axes
of the variables are interchanged, the equation
The equation E = ( IR r+ ) can becomes:
be used to determine the internal
resistance of a cell and the e.m.f 1 = 1 R + r
of the cell experimentally. From I E E
Figure 2.19, different R () Therefore, the slope of this graph (Figure 2.21)
values can be used to give represents the reciprocal of the cell’s e.m.f, and
different values of current, I (A), the vertical intercept represents the product of the
and the two sets of values are slope and the cell’s internal resistance.

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Physics Form 2 Final.indd 51 25/10/2025 10:26