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Questions on EMF and internal resistance

Q2.

A very high resistance voltmeter reads 20V when connected across the terminals of a d.c. power supply. The high resistance meter is disconnected and a second voltmeter of resistance 1.0k is then connected across the supply. The second meter gives a reading of 16V.

(i) State the e.m.f. of the power supply.

20V

(ii) Calculate the current which flows through the second meter.

V = IR
I = V/R = 16/1000
= 1.6 × 10–2 A


(iii) Calculate the internal resistance of the power supply.

'lost volts' = Ir
20 - 16 = 1.6 × 10–2 × r s
r = 250 s s
OR
EMF/I = R + r
20/1.6 × 10–2 =1000 + r s
r = 250 s s
(iv) Show that the current is equal to 0.080A when the supply is short circuited.
Short circuited - no external resistance
EMF = Ir
I = 20/250 = 8.0 × 10–2 A

(Total 5 marks)