Current Electricity
Physics · 50 practice questions
E.M.F., potential difference, current, internal resistance and lost volts; Ohm's law, resistivity and conductivity; measurement of resistance, the meter bridge; resistors in series and parallel; the potentiometer method of measuring e.m.f., current and internal resistance; electrical networks (Kirchhoff's laws).
- Differentiate between e.m.f., p.d., current and internal resistance.
- Apply Ohm's law and use the metre bridge to calculate resistance.
- Compute effective resistance of series/parallel arrangements and determine resistivity/conductivity.
- Measure e.m.f., current and internal resistance using a potentiometer, and apply Kirchhoff's laws to electrical networks.
Try a few questions
- Practice question 1
Ohm's law states that, at constant temperature, the potential difference across a conductor is
- A.inversely proportional to the current flowing through it
- B.independent of the current flowing through it
- C.directly proportional to the square of the current flowing through it
- D.directly proportional to the current flowing through it
Show the answer
D. directly proportional to the current flowing through it
This is the standard statement of Ohm's law, V = IR, at constant temperature.
- Practice question 2
A battery of e.m.f. 6V drives a current of 0.4A through a circuit with a total resistance of 15Ω. This is consistent with Ohm's law since
- A.V = IR = 0.4 x 15 = 60V, which does not match the given e.m.f.
- B.V = I/R = 0.4/15, which gives an entirely different, inconsistent value
- C.Ohm's law cannot be applied to a circuit with a battery involved
- D.V = IR = 0.4 x 15 = 6V, matching the given e.m.f.
Show the answer
D. V = IR = 0.4 x 15 = 6V, matching the given e.m.f.
Since IR exactly equals the stated 6V e.m.f., this confirms the given values are consistent with Ohm's law.
- Practice question 3
A current of 2A flows through a resistor when a potential difference of 8V is applied. The resistance of the resistor is
- A.16 Ω
- B.6 Ω
- C.4 Ω
- D.0.25 Ω
Show the answer
C. 4 Ω
Resistance = voltage ÷ current = 8 ÷ 2 = 4 Ω.
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