Topic Menu

Cyber-chess a chess instruction site

 

Questions on EMF and internal resistance

Q8. In the circuit shown below, the battery, of emf 6.0V, has negligible internal resistance.

(a) Calculate the current through the ammeter when the switch S is

(i) open,

R = 20 + 60= 80

V = IR so I =V/R

I = 6.0/80 = 0 075 A

(ii) closed

When the switch is closed the 60 resistor is 'shorted out'. It is in effect a resistor in parallel with a piece of connecting wire. The 'short' wire and resistance parallel combination have less resistance than the wire itself - therefore can be ignored! It is as if there is just a piece of connecting wire at that point in the circuit.

R = 20

I =V/R = 6.0/20 = 0.3 A

(3 MAX)

(b) The switch S is now replaced with a voltmeter of infinite resistance. Determine the reading on the voltmeter.

The voltmeter has such a high resistance that the parallel arrangement it makes with the resistor has a value of the resistor alone (remember that the reciprocal of infinity is zero, so using the 'resistors in parallel' equation, you get that result.

We now have the same circuit as in (a)(i), but with the addition of the voltmeter.

The voltmeter will have the same voltage across it as the 60 ,

V = IR = 0.075 × 60 = 4.5 V

OR the voltmeter will read 60/(20 + 60) = 3/4 of the total voltage

3/4 x 6 = 4.5 V

(2 MAX)

(Total 5 marks)