Page 59 - Physics_Form_2

Current electricity

From (a) Plot a graph of V against I
E = ( IR r+ ) (b) Determine the value of
resistance R
E = 0.6E Ir+ 3. Study Figure 2.23 and the given
Ir  E  0.6E instructions to answer the questions
FOR ONLINE READING ONLY
Ir = 0.4E (1) that follow.
From Ohm’s law
V = IR r

3I = 0.6E
0.6E (2) S
I =
3

Substituting Equation (2) into (1) and R
solving for ,r you get r  2  . Figure 2.23

Exercise 2.1 Switch Ammeter Voltmeter
condition reading reading
1. A student measures the voltage of a (mA) (V)
cell in two scenarios. In scenario 1, Open 0 9
he connects the voltmeter directly to
the cell and records a value of 1.5 V. Closed 250 7
In scenario 2, he adds an unknown (a) Why does the reading on the
resistor to the circuit and records a voltmeter drop when the switch
voltage of 1.32 V. is closed?
(a) What conclusions can be drawn (b) Calculate the unknown external
from these results? load resistance R.
(b) Explain the terms associated (c) Calculate the internal resistance
with the values recorded in
scenarios 1 and 2. 4. A student sets up the schematic
2. The following results were obtained diagram as in Figure 2.24 to
investigate the internal resistance
in an experiment to determine the of a cell where R  is a fixed
2

value of resistance. resistor, r is the internal resistance
Voltage 0.04 0.08 0.2 0.21 0.22 of the cell, A is an ammeter and V is
(mV) a voltmeter across the cell. When the

Current 5.5 10.5 20.3 28.9 30 switch is closed, the ammeter reads
(mA) 0.4 A, the voltmeter reads 1.2 V. If
the emf, E of the cell is known to
From the experimental results, be 1.6 V:
using ICT tools or otherwise: (a) Draw the circuit diagram clearly

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Physics Form 2 Final.indd 53 25/10/2025 10:26