Electrical experiments: circuits, meters and resistance

In this note

Rules to learn

Circuit symbols

A circuit diagram mark needs every symbol right. These are the ones that come up:

Connecting and reading meters

Read meters with the method in Measurements and readings: value of one division, count from the label below the needle, half-division. An ammeter that reads off the end of its scale (beyond full-scale deflection) needs a larger range or a smaller current.

Resistance experiments

Precautions and faults

From the past papers

Past paper 1

How to start: Draw the loop first (supply, switch, LDR, resistor), then the voltmeter across the LDR only. V1 is a meter reading (0.80 V); I comes from the OTHER voltmeter reading across the 560 Ω resistor.

✓ Show the answer have a go on paper first

1. (a) One series loop: power supply, switch, LDR and the 560 Ω resistor, every symbol correct and nothing else.
2. (b)(i) Voltmeter across the LDR (parallel).
4. (b)(ii) \(V_1=0.80\) V. (iii) To stop the battery running down / the components heating up.
5. (c)(i) \(I=\dfrac{2.18}{560}=3.89\times10^{-3}\) A. (ii) \(R_{LDR}=\dfrac{0.80}{0.00389}=206\ \Omega\).
6. (d) V across the LDR is larger in the dark (1.94 V > 0.80 V): the resistance of the LDR increases as the light decreases.
7. (e) The reading stays about the same while the card is far away, then increases as the card gets close to the LDR.

Part	Answer / working	Marks	Guidance
(a)	(any) series circuit drawn	B1
	power supply, switch, LDR, resistor and no other components, in series, all symbols correct	B1
(b)(i)	voltmeter connected in parallel across the LDR, symbol correct	B1
(b)(ii)	0.8(0) (V)	B1
(b)(iii)	to stop the battery running down or the resistor / circuit / components heating up	B1
(c)(i)	3.89 × 10⁻³ / 0.0039 (A)	B1
(c)(ii)	205 / 206 / 210 (Ω)	B1
(d)	resistance of the LDR increases	B1
(e)	reading initially (more or less) constant as the distance decreases	B1
	(then) increases as the card gets close to the LDR	B1
Total: 10 marks

Examiner tip (b)(iii) is the switch-off precaution in disguise: heating and battery life.

Past paper 2

How to start: Read both meters first (count the divisions). Then every row is R = V/I or I = V/R. Part (d) wants each change linked to the current going DOWN.

✓ Show the answer have a go on paper first

1. (a) \(V=0.84\) V, \(I=0.36\) A.
2. (b) \(R=0.84\div0.36=2.33\to\) 2.3 Ω (2 s.f., like the readings).
3. (c) 13 Ω row: \(R=0.83\div0.27=3.1\ \Omega\). 20 Ω row: \(I=0.81\div4.5=0.18\) A.
4. (d)(i) As the current decreases, the p.d. across the diode stays (fairly) constant / decreases slightly. (ii) As the current decreases, the resistance of the diode increases.
5. (e) The diode (or the power supply) was connected the wrong way round.
6. (f) A variable resistor (rheostat) - the rectangle with an arrow through it, not the thermistor symbol.

Part	Answer / working	Marks	Guidance
(a)	V = 0.84 (V)	B1
	I = 0.36 (A)	B1
(b)	2.3(3…)	B1
	2.3 (Ω)	B1
(c)	3.1 (Ω)	B1
	0.18 (A)	B1
(d)(i)	as the current decreases the p.d. across the diode remains (fairly) constant / decreases (slightly)	B1
(d)(ii)	as the current decreases the resistance increases	B1
(e)	diode / power supply / ammeter connected the wrong way round	B1
(f)	variable resistor and correct symbol	B1
Total: 10 marks

Examiner tip The report: "directly / inversely proportional" earned nothing in (d) - describe each change linked to the current.

Mistakes to avoid

• ✗ thermistor symbol for a variable resistor✓ rectangle with an arrow through it
• ✗ voltmeter in the main loop✓ voltmeter across the component
• ✗ "the switch was not closed" for a complete circuit that fails✓ "the diode was connected the wrong way round"Read what the question says is true.
• ✗ R = I/V✓ R = V/I
• ✗ "inversely proportional" as a description✓ "as the current decreases, R increases"

Summary

Practice

Can I do this?

Quiz