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Questions on EMF and internal resistance

Q1.

A battery is connected to a 10 resistor as shown. The e.m.f. (electromotive force) of the battery is 12V.

(a)

(i) Explain what is meant by the e.m.f. of a battery.

The EMF is the energy provided by a battery per unit charge
(OR when no current flows it is the potential difference across the battery )

(ii) When the switch is open the voltmeter reads 12.0V and when it is closed it reads 11.5 V. Explain why the readings are different.

A potential difference is developed across the internal resistance

(3)

(b) Calculate the internal resistance of the battery.

I =11.5/10 = 1.15A

'lost volts' = Ir
12 – 11.5 = 1.15 × r
r = 0.43

OR

EMF/I = R + r

12/1.15 =10 + r

10.43 - 10 = r = 0.43

(3)
(Total 6 mark)

The EMF is the energy provided by a battery per unit charge
(OR when no current flows it is the potential difference across the battery )

A potential difference is developed across the internal resistance

I =11.5/10 = 1.15A

'lost volts' = Ir
12 – 11.5 = 1.15 × r
r = 0.43

OR

EMF/I = R + r

12/1.15 =10 + r

10.43 - 10 = r = 0.43

Topic Menu

The EMF is the energy provided by a battery per unit charge (OR when no current flows it is the potential difference across the battery )

A potential difference is developed across the internal resistance

I =11.5/10 = 1.15A

'lost volts' = Ir 12 – 11.5 = 1.15 × r r = 0.43

OR

EMF/I = R + r

12/1.15 =10 + r

10.43 - 10 = r = 0.43

Follow me...

The EMF is the energy provided by a battery per unit charge
(OR when no current flows it is the potential difference across the battery )

'lost volts' = Ir
12 – 11.5 = 1.15 × r
r = 0.43