Physics 9702 · Paper 2

Effect-of-change Questions

One thing in a circuit is changed, and you must state and explain what happens to a reading.
The marks are for a chain of short steps. Miss one link and you lose its mark.

The chain you must write

Start from what changed: total resistance up or down.
Then the current, from $I = E/(R + r)$ : resistance down means current up.
Then the lost volts $Ir$ : more current means more lost volts.
Then the terminal p.d. $V = E - Ir$ : more lost volts means lower terminal p.d.
End at the reading the question asked about.
Two things never change: E and r. They belong to the battery, not to the circuit. Writing “E and r do not change” is often one of the marks.

Write the chain link by link. “The current changes so the voltage changes” has no links in it, so it scores nothing.

Adding a parallel branch

Adding a resistor in parallel gives the current an extra path, so the total resistance falls. Many students guess the opposite because “a resistor was added”.

Worked example

A cell, two resistors, then a third in parallel

A cell of e.m.f. 1.50 V and internal resistance r drives resistors of 10 Ω and 15 Ω in series. The terminal p.d. is 1.35 V. (a) Find r. (b) A third resistor is now connected in parallel with the pair. State and explain what happens to the terminal p.d.

(a) Current: $I = \dfrac{1.35}{10 + 15} = 0.054$ A.
$r = \dfrac{E - V}{I} = \dfrac{0.15}{0.054} = 2.8 \; \Omega$ .
(b) The chain: total resistance falls (extra parallel path), e.m.f. unchanged, so the current in the cell rises.
More current means larger lost volts $Ir$ , so the terminal p.d. $E - Ir$ decreases.

Answer

r = 2.8 \; \Omega; \; \text{terminal p.d. falls}

(9702/21/M/J/24 Q6)

Common mistake

“A resistor was added, so the resistance goes up.” Not in parallel. In parallel the resistance always goes down, because the current gets another path. Check series or parallel before you write the first link.

Two more changes the exam asks about

A lamp warms up. When a switch sends current through a lamp, its filament heats up and its resistance rises (from the Electricity chapter). So the current falls a little just after switching on. (9702/22/F/M/24 Q7)
Connecting a voltmeter is itself a change. A real voltmeter is a resistor in parallel, so it lowers the resistance of whatever it is connected across. This is the worked example that the meters page mentioned:

Worked example

What a real voltmeter does to a reading

A battery of negligible internal resistance drives a 100 Ω and a 400 Ω resistor in series. A voltmeter of resistance 1200 Ω across the 400 Ω resistor reads 9.0 V. (a) Find the e.m.f. (b) Show that a 12 kΩ voltmeter would read about 9.5 V.

(a) The voltmeter is in parallel with the 400 Ω: $\dfrac{400 \times 1200}{400 + 1200} = 300 \; \Omega$ .
That 300 Ω carries 9.0 V, so $I = 9.0/300 = 0.030$ A and $E = 0.030 \times (100 + 300) = 12$ V.
(b) With 12 kΩ: $\dfrac{400 \times 12000}{12400} = 387 \; \Omega$ , so the reading is $12 \times \dfrac{387}{100 + 387} = 9.5$ V.
With no voltmeter at all, the 400 Ω would take $12 \times \dfrac{400}{500} = 9.6$ V. The larger the voltmeter's resistance, the closer the reading gets to the true value.

Answer

E = 12 \text{ V}; \; 9.5 \text{ V} \to 9.6 \text{ V}

This is why the ideal voltmeter has infinite resistance: it is the only voltmeter that does not change the p.d. it is trying to read.

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Effect-of-change Questions

One thing in a circuit is changed, and you must state and explain what happens to a reading.
The marks are for a chain of short steps. Miss one link and you lose its mark.

The chain you must write

Start from what changed: total resistance up or down.
Then the current, from $I = E/(R + r)$ : resistance down means current up.
Then the lost volts $Ir$ : more current means more lost volts.
Then the terminal p.d. $V = E - Ir$ : more lost volts means lower terminal p.d.
End at the reading the question asked about.
Two things never change: E and r. They belong to the battery, not to the circuit. Writing “E and r do not change” is often one of the marks.

Write the chain link by link. “The current changes so the voltage changes” has no links in it, so it scores nothing.

Adding a parallel branch

Adding a resistor in parallel gives the current an extra path, so the total resistance falls. Many students guess the opposite because “a resistor was added”.

Worked example

A cell, two resistors, then a third in parallel

(a) Current: $I = \dfrac{1.35}{10 + 15} = 0.054$ A.
$r = \dfrac{E - V}{I} = \dfrac{0.15}{0.054} = 2.8 \; \Omega$ .
(b) The chain: total resistance falls (extra parallel path), e.m.f. unchanged, so the current in the cell rises.
More current means larger lost volts $Ir$ , so the terminal p.d. $E - Ir$ decreases.

Answer

r = 2.8 \; \Omega; \; \text{terminal p.d. falls}

(9702/21/M/J/24 Q6)

Common mistake

Two more changes the exam asks about

A lamp warms up. When a switch sends current through a lamp, its filament heats up and its resistance rises (from the Electricity chapter). So the current falls a little just after switching on. (9702/22/F/M/24 Q7)
Connecting a voltmeter is itself a change. A real voltmeter is a resistor in parallel, so it lowers the resistance of whatever it is connected across. This is the worked example that the meters page mentioned:

Worked example

What a real voltmeter does to a reading

(a) The voltmeter is in parallel with the 400 Ω: $\dfrac{400 \times 1200}{400 + 1200} = 300 \; \Omega$ .
That 300 Ω carries 9.0 V, so $I = 9.0/300 = 0.030$ A and $E = 0.030 \times (100 + 300) = 12$ V.
(b) With 12 kΩ: $\dfrac{400 \times 12000}{12400} = 387 \; \Omega$ , so the reading is $12 \times \dfrac{387}{100 + 387} = 9.5$ V.
With no voltmeter at all, the 400 Ω would take $12 \times \dfrac{400}{500} = 9.6$ V. The larger the voltmeter's resistance, the closer the reading gets to the true value.

Answer

E = 12 \text{ V}; \; 9.5 \text{ V} \to 9.6 \text{ V}

This is why the ideal voltmeter has infinite resistance: it is the only voltmeter that does not change the p.d. it is trying to read.