'A' Level Questions - The Oscilloscope
Q4.
(a) The circuit shown in the diagram on the left of the above, may be used to determine the internal resistance of a battery. An oscilloscope is connected across the battery as shown. The diagram on the right (above) represents the screen of the oscilloscope.
The time base of the oscilloscope is switched off throughout the experiment. Initially the switches S1 and S2 are both open. Under these conditions the spot on the oscilloscope screen is at A.
(i) Switch S1 is now closed, with S2 remaining open. The spot moves to B. State what the deflection AB represents.
AB represents the emf (of the battery) 
(ii) Switch S1 is kept closed and S2 is also closed. The spot moves to C. State what the deflection AC represents.
AC represents the voltage across the battery when current flows [or terminal voltage or pd supplied to the circuit]
(iii) The vertical sensitivity of the oscilloscope is 0.50 V div–1. Calculate the current through the 14 
resistor with both switches closed.
V = IR
V = 3 divisions = 1.5 volts
I = V/R = 1.5/14 = 0.107
I = 0.11 A
(iv) Hence, calculate the internal resistance of the battery.
Method 1
Lost volts = EMF - V = Ir
Lost volts = 3.5 x 0.5 = 1.75 volts
r = (1.75 - 1.5) /I = 0.25/0.107 = 2.33
r = 2.3
Method 2
EMF = I(R + r)
1.75 = 0.105(14 +r)
1.75/0.107 = 14 + r
16.4 = 14 + r
r = 2.4
(6 marks)
(b) The oscilloscope is now connected to an alternating voltage source of rms value 3.5 V.
(i) Calculate the peak value of the alternating voltage.
peak value = 3.5 x 
= 4.9 V
(ii) Draw on the diagram below what you would expect to see on the oscilloscope screen, if the time base is still switched off and the voltage sensitivity is altered to 2.0 V div–1.
Marks awarded for:
- vertical line of height about 4.9 divisions in length and
- centred on the central axis
(3 marks)
(Total 9 marks)
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