2.3 Potential difference and resistance
9 exam questions worth 21 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
0 / 21
The current in a 4 Ω resistor is 3 A. Calculate the potential difference across the resistor.
Mark scheme
V = I R = 3 × 4 (1); 12 V (1)
Model answer
V = I R = 3 × 4 = 12 V.
A lamp has a potential difference of 12 V across it and a current of 0.5 A through it. Calculate the resistance of the lamp.
Mark scheme
R = V / I (1); 12 / 0.5 (1); 24 Ω (1)
Model answer
R = V / I = 12 / 0.5 = 24 Ω.
A 9 V battery transfers 40 C of charge through a circuit. Calculate the energy transferred.
Mark scheme
E = Q V = 40 × 9 (1); 360 J (1)
Model answer
E = Q V = 40 × 9 = 360 J.
State how a voltmeter must be connected to measure the potential difference across a component, and what it measures.
Mark scheme
in parallel with / across the component (1); the energy transferred per unit charge / potential difference between two points (1)
Model answer
A voltmeter is connected in parallel with the component. It measures the potential difference across it, which is the energy transferred per coulomb of charge.
A student increases the resistance of the variable resistor in a circuit. The potential difference of the cell stays the same. Explain what happens to the current.
Mark scheme
the total resistance of the circuit increases (1); so the current decreases (as I = V / R) (1)
Model answer
The total resistance of the circuit increases. Because the potential difference of the cell is constant, the current decreases (I = V / R).
Which of these is the same as one volt?
Choose one answer.
Correct Not quite: the answer is One joule per coulomb
Mark scheme
A: one joule per coulomb
Model answer
One joule per coulomb (A), because potential difference is energy transferred per unit charge.
A 6.0 V battery drives a current of 0.50 A through a lamp for 2.0 minutes. Calculate the charge that flows and the energy transferred.
Mark scheme
t = 120 s (1); Q = I t = 0.50 × 120 = 60 C (1); E = Q V = 60 × 6.0 (1); 360 J (1)
Model answer
t = 120 s. Q = I t = 60 C. E = Q V = 60 × 6.0 = 360 J.
A current of 25 mA flows through a resistor of resistance 400 Ω. Calculate the potential difference across the resistor.
Mark scheme
25 mA = 0.025 A (1); V = I R = 0.025 × 400 (1); 10 V (1)
Model answer
I = 0.025 A. V = I R = 0.025 × 400 = 10 V.
A student says that 'the voltage goes through the lamp'. Explain what is wrong with this statement.
Mark scheme
potential difference is measured across a component (1); current is what flows through it (1)
Model answer
Potential difference is a difference between two points, so it is across a component. It is the current (the flow of charge) that goes through the lamp.
Found a mistake?