Circuits - Series and Parallel
Q3. In the circuit shown below, the battery provides a potential difference of 6.0V.
![](../EMF/7.png)
(a) Calculate the current through the ammeter when the switch S is
(i) open,
![](series.png)
R = 20 + 60= 80 ![](../../../../graphics/symbols/omega_red.png)
![](../../../../graphics/symbols/nuclides/ticksmall.png)
V = IR so I =V/R
I = 6.0/80 = 0 075 A ![](../../../../graphics/symbols/nuclides/ticksmall.png)
(ii) closed
R = 20
![](../../../../graphics/symbols/nuclides/ticksmall.png)
I =V/R = 6.0/20 = 0.3 A ![](../../../../graphics/symbols/nuclides/ticksmall.png)
(3 MAX)
(b) The switch S is now replaced with a voltmeter of infinite resistance. Determine the reading on the voltmeter.
The voltmeter will have the same voltage across it as the 60
, ![](../../../../graphics/symbols/nuclides/ticksmall.png)
V = IR = 0.075 × 60 = 4.5 V
OR the voltmeter will read 60/(20 + 60) = 3/4 of the total voltage ![](../../../../graphics/symbols/nuclides/ticksmall.png)
3/4 x 6 = 4.5 V ![](../../../../graphics/symbols/nuclides/ticksmall.png)
(2 MAX)
(Total 5 marks)