AP Physics C: Electricity and Magnetism Practice Test 16 – 60 Questions with Answers

Test yourself with 60 original AP Physics C: Electricity and Magnetism practice questions covering all 6 units of the course: electric charges, fields, and Gauss’s law, electric potential, conductors and capacitors, electric circuits, magnetic fields and electromagnetism, and electromagnetic induction. Every answer includes the full working. All questions are calculus-based.

Unit 8: Electric Charges, Fields, and Gauss’s Law

Q1. Two point charges carry charges of 10 μC and 2 μC and are held 0.1 m apart in air. What is the magnitude of the electrostatic force between them?

  • (A) 36 N
  • (B) 9 N
  • (C) 27 N
  • (D) 18 N
Show Answer

Answer: D – Using Coulomb’s law, F = kq1q2/r² = (9.00×109)(10×10-6)(2×10-6)/(0.1)² = 18 N.

Q2. A point charge of 6 μC is fixed in vacuum. What is the magnitude of the electric field at a point 0.8 m from the charge?

  • (A) 84.4 kN/C
  • (B) 169 kN/C
  • (C) 42.2 kN/C
  • (D) 127 kN/C
Show Answer

Answer: A – E = kQ/r² = (9.00×109)(6×10-6)/(0.8)² = 8.44×104 N/C = 84.4 kN/C.

Q3. A closed Gaussian surface encloses a net charge of 4 nC. What is the net electric flux through the surface?

  • (A) 904 N·m²/C
  • (B) 226 N·m²/C
  • (C) 678 N·m²/C
  • (D) 452 N·m²/C
Show Answer

Answer: D – By Gauss’s law, Φ = Qenc/ε0 = (4×10-9)/(8.85×10-12) = 452 N·m²/C.

Q4. An infinitely long straight wire carries a uniform linear charge density of 0.4 μC/m. What is the magnitude of the electric field at a perpendicular distance of 0.3 m from the wire?

  • (A) 48 kN/C
  • (B) 12 kN/C
  • (C) 36 kN/C
  • (D) 24 kN/C
Show Answer

Answer: D – For an infinite line, E = 2kλ/r = 2(9.00×109)(0.4×10-6)/(0.3) = 24 kN/C.

Q5. A very large flat sheet carries a uniform surface charge density of 4 μC/m². What is the magnitude of the electric field just outside the sheet?

  • (A) 452 kN/C
  • (B) 226 kN/C
  • (C) 113 kN/C
  • (D) 339 kN/C
Show Answer

Answer: B – For an infinite sheet, E = σ/(2ε0) = (4×10-6)/(2×8.85×10-12) = 226 kN/C.

Q6. A solid insulating sphere of radius 0.25 m carries a total charge of 3 μC uniformly distributed through its volume. What is the magnitude of the electric field at a distance of 0.15 m from its centre?

  • (A) 518 kN/C
  • (B) 130 kN/C
  • (C) 259 kN/C
  • (D) 389 kN/C
Show Answer

Answer: C – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R³ = (9.00×109)(3×10-6)(0.15)/(0.25)³ = 259 kN/C.

Q7. A thin spherical shell of radius 0.25 m carries a total charge of 2 μC uniformly on its surface. What is the magnitude of the electric field at a distance of 0.6 m from the centre of the shell?

  • (A) 100 kN/C
  • (B) 25 kN/C
  • (C) 75 kN/C
  • (D) 50 kN/C
Show Answer

Answer: D – Outside the shell the field is that of a point charge at the centre: E = kQ/r² = (9.00×109)(2×10-6)/(0.6)² = 50 kN/C.

Q8. A thin rod of length 0.5 m carries a total charge of 4 μC uniformly along its length. Using integration, what is the magnitude of the electric field at a point on the rod’s axis at a distance of 0.1 m from its nearer end?

  • (A) 1.2×103 kN/C
  • (B) 600 kN/C
  • (C) 300 kN/C
  • (D) 900 kN/C
Show Answer

Answer: B – Integrating dE = kλdx/x² from x = 0.1 m to 0.6 m gives E = kQ/[a(a + L)] = (9.00×109)(4×10-6)/[(0.1)(0.6)] = 600 kN/C.

Q9. A thin ring of radius 0.15 m carries a total charge of 2 μC uniformly around its circumference. What is the magnitude of the electric field on the ring’s axis at a distance of 0.5 m from its centre?

  • (A) 127 kN/C
  • (B) 63.3 kN/C
  • (C) 31.6 kN/C
  • (D) 94.9 kN/C
Show Answer

Answer: B – On the axis, E = kQz/(z² + R²)3/2 = (9.00×109)(2×10-6)(0.5)/(0.5² + 0.15²)3/2 = 63.3 kN/C.

Q10. An electric dipole consists of charges of +4 μC and -4 μC separated by 0.08 m. What is the approximate magnitude of the electric field on the dipole axis at a distance of 0.4 m from its centre, where the distance is much larger than the separation?

  • (A) 90 kN/C
  • (B) 180 kN/C
  • (C) 45 kN/C
  • (D) 135 kN/C
Show Answer

Answer: A – On the axis far from a dipole, E ≈ 2kp/r³ with p = qd = (4×10-6)(0.08) C·m, giving E = 90 kN/C.

Q11. A closed surface encloses a dipole consisting of equal and opposite charges, with no other charge inside. What is the net electric flux through the surface, and why?

  • (A) Zero, because the enclosed net charge is zero
  • (B) Zero, because the field is zero everywhere on the surface
  • (C) Non-zero, because each charge produces flux
  • (D) Non-zero, because the dipole field never cancels
Show Answer

Answer: A – Gauss’s law gives Φ = Qenc/ε0. The dipole contributes +q and -q, so Qenc = 0 and the net flux is zero, even though the field on the surface is not zero.

Q12. Two positive point charges of 3 μC and 1 μC lie on the x-axis. A point P lies on the same axis beyond both charges, 0.6 m from the first charge and 0.3 m from the second charge. What is the magnitude of the net electric field at P?

  • (A) 175 kN/C
  • (B) 350 kN/C
  • (C) 87.5 kN/C
  • (D) 262 kN/C
Show Answer

Answer: A – Both fields point away from their charges, in the same direction at P, so E = kq1/r1² + kq2/r2² = 175 kN/C.

Unit 9: Electric Potential

Q13. A point charge of 5 μC is fixed in vacuum. What is the electric potential at a distance of 0.4 m from the charge, taking the potential at infinity to be zero?

  • (A) 225 kV
  • (B) 56.2 kV
  • (C) 112 kV
  • (D) 169 kV
Show Answer

Answer: C – V = kQ/r = (9.00×109)(5×10-6)/(0.4) = 112 kV.

Q14. Two point charges of 2 μC and 6 μC are held 0.2 m apart. What is the electric potential energy of this pair, taking zero energy at infinite separation?

  • (A) 1.08 J
  • (B) 0.27 J
  • (C) 0.54 J
  • (D) 0.81 J
Show Answer

Answer: C – U = kq1q2/r = (9.00×109)(2×10-6)(6×10-6)/(0.2) = 0.54 J.

Q15. A thin ring of radius 0.2 m carries a total charge of 5 μC uniformly around its circumference. What is the electric potential on the ring’s axis at a distance of 0.4 m from its centre?

  • (A) 201 kV
  • (B) 50.3 kV
  • (C) 101 kV
  • (D) 151 kV
Show Answer

Answer: C – Every element of charge is at the same distance √(z² + R²) from the point, so V = kQ/√(z² + R²) = 101 kV.

Q16. A thin rod of length 0.5 m carries a uniform linear charge density of 1 μC/m. By integrating the contributions of its charge elements, what is the electric potential at a point on the rod’s axis 0.2 m from its nearer end, with zero potential at infinity?

  • (A) 22.5 kV
  • (B) 11.3 kV
  • (C) 5.64 kV
  • (D) 16.9 kV
Show Answer

Answer: B – V = ∪kλdx/x = kλ ln[(a + L)/a] = (9.00×109)(1×10-6) ln(0.7/0.2) = 11.3 kV.

Q17. A uniform electric field of 400 N/C points in the +x direction. A charge is moved 0.1 m in the +x direction. By how much does the electric potential change?

  • (A) 40 V
  • (B) -80 V
  • (C) -20 V
  • (D) -40 V
Show Answer

Answer: D – ΔV = -EΔx = -(400)(0.1) = -40 V. The potential falls in the direction of the field.

Q18. The electric potential along the x-axis is V(x) = 50x² + 20x volts, where x is in metres. What is the x-component of the electric field at x = 2 m?

  • (A) 220 V/m
  • (B) -440 V/m
  • (C) -220 V/m
  • (D) -110 V/m
Show Answer

Answer: C – Ex = -dV/dx = -(100x + 20). At x = 2 m this gives Ex = -220 V/m.

Q19. A charge of 3 μC is moved through a potential increase of 5000 V by an external agent at constant speed. How much work does the external agent do?

  • (A) 15 mJ
  • (B) 30 mJ
  • (C) 7.5 mJ
  • (D) 22.5 mJ
Show Answer

Answer: A – W = qΔV = (3×10-6)(5000) = 15 mJ.

Q20. A conducting sphere of radius 0.3 m carries a total charge of 1 μC. What is the electric potential at the surface of the sphere, with zero potential at infinity?

  • (A) 60 kV
  • (B) 30 kV
  • (C) 15 kV
  • (D) 45 kV
Show Answer

Answer: B – Outside a charged sphere V = kQ/r, so at the surface V = kQ/R = (9.00×109)(1×10-6)/(0.3) = 30 kV.

Q21. At a certain point in space the electric potential is zero. Which statement about the electric field at that point is necessarily correct?

  • (A) The field may be zero or non-zero; zero potential does not fix the field
  • (B) The field must be zero
  • (C) The field must be at its maximum value
  • (D) The field must point toward the point
Show Answer

Answer: A – Potential is defined up to an additive constant and depends on the reference point, while the field is the negative gradient of the potential. A single point of zero potential says nothing by itself about the gradient there.

Unit 10: Conductors and Capacitors

Q22. A parallel-plate capacitor has plates of area 100 cm² separated by 2 mm of air. What is its capacitance?

  • (A) 88.5 pF
  • (B) 22.1 pF
  • (C) 44.2 pF
  • (D) 66.4 pF
Show Answer

Answer: C – C = ε0A/d = (8.85×10-12)(100×10-4)/(2×10-3) = 44.2 pF.

Q23. An air-filled capacitor has capacitance 60 pF. It is completely filled with a dielectric of dielectric constant 5 while its geometry is unchanged. What is the new capacitance?

  • (A) 600 pF
  • (B) 300 pF
  • (C) 150 pF
  • (D) 450 pF
Show Answer

Answer: B – C = κC0 = (5)(60 pF) = 300 pF.

Q24. A capacitor of capacitance 4 μF is charged to 200 V. How much energy is stored in the capacitor?

  • (A) 160 mJ
  • (B) 40 mJ
  • (C) 120 mJ
  • (D) 80 mJ
Show Answer

Answer: D – U = (1/2)CV² = (1/2)(4×10-6)(200)² = 80 mJ.

Q25. Capacitors of 8 μF and 4 μF are connected in series. What is their equivalent capacitance?

  • (A) 5.33 μF
  • (B) 1.33 μF
  • (C) 2.67 μF
  • (D) 4 μF
Show Answer

Answer: C – 1/Ceq = 1/8 + 1/4, so Ceq = 8×4/(8 + 4) = 2.67 μF.

Q26. Capacitors of 4 μF, 4 μF and 2 μF are connected in parallel. What is their equivalent capacitance?

  • (A) 20 μF
  • (B) 10 μF
  • (C) 5 μF
  • (D) 15 μF
Show Answer

Answer: B – For a parallel combination, Ceq = C1 + C2 + C3 = 10 μF.

Q27. A capacitor of capacitance 6 μF is connected across a 20 V battery. What charge is stored on each plate?

  • (A) 240 μC
  • (B) 60 μC
  • (C) 180 μC
  • (D) 120 μC
Show Answer

Answer: D – Q = CV = (6 μF)(20 V) = 120 μC.

Q28. A parallel-plate capacitor has a potential difference of 500 V across plates separated by 4 mm. What is the magnitude of the uniform electric field between the plates, ignoring edge effects?

  • (A) 250 kV/m
  • (B) 62.5 kV/m
  • (C) 125 kV/m
  • (D) 188 kV/m
Show Answer

Answer: C – E = V/d = (500)/(4×10-3) = 125 kV/m.

Q29. A spherical capacitor consists of concentric conducting spheres of radii 0.08 m and 0.12 m. What is its capacitance?

  • (A) 26.7 pF
  • (B) 53.4 pF
  • (C) 13.3 pF
  • (D) 40 pF
Show Answer

Answer: A – C = 4πε0ab/(b – a) = (4π×8.85×10-12)(0.08)(0.12)/(0.12 – 0.08) = 26.7 pF.

Unit 11: Electric Circuits

Q30. A resistor of 8 Ω is connected across a 9 V battery of negligible internal resistance. What is the current through the resistor?

  • (A) 2.25 A
  • (B) 0.562 A
  • (C) 1.69 A
  • (D) 1.12 A
Show Answer

Answer: D – I = V/R = 9/8 = 1.12 A.


Q31. Resistors of 3 Ω, 6 Ω and 5 Ω are connected in series across a 12 V battery. What is the current in the circuit?

  • (A) 1.71 A
  • (B) 0.429 A
  • (C) 1.29 A
  • (D) 0.857 A
Show Answer

Answer: D – Req = 3 + 6 + 5 = 14 Ω, so I = V/Req = 12/14 = 0.857 A.

Q32. Resistors of 6 Ω and 4 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 4.8 Ω
  • (B) 1.2 Ω
  • (C) 3.6 Ω
  • (D) 2.4 Ω
Show Answer

Answer: D – 1/Req = 1/6 + 1/4, so Req = 6×4/(6 + 4) = 2.4 Ω.

Q33. A 6 Ω resistor is connected across a 20 V supply. At what rate is electrical energy converted to heat in the resistor?

  • (A) 133 W
  • (B) 33.3 W
  • (C) 66.7 W
  • (D) 100 W
Show Answer

Answer: C – P = V²/R = (20)²/6 = 66.7 W.

Q34. In a single loop, two batteries of emf 18 V and 9 V oppose each other, and the total resistance in the loop is 9 Ω. What is the current in the loop?

  • (A) 2 A
  • (B) 0.5 A
  • (C) 1 A
  • (D) 1.5 A
Show Answer

Answer: C – Applying Kirchhoff’s loop rule, (18 – 9) = I(9), so I = 9/9 = 1 A, in the direction of the larger emf.

Q35. A battery of emf 15 V and internal resistance 2 Ω is connected to an external resistor of 17 Ω. What is the terminal voltage of the battery?

  • (A) 26.8 V
  • (B) 13.4 V
  • (C) 6.71 V
  • (D) 20.1 V
Show Answer

Answer: B – I = E/(R + r) = 15/(17 + 2) A, so V = E – Ir = 13.4 V.

Q36. A resistor of 20 kΩ and a capacitor of 20 μF are connected in series. What is the time constant of this RC circuit?

  • (A) 400 ms
  • (B) 800 ms
  • (C) 200 ms
  • (D) 600 ms
Show Answer

Answer: A – τ = RC = (20×103 Ω)(20×10-6 F) = 400 ms.

Q37. An uncharged capacitor is connected in series with a 2 kΩ resistor across a 15 V battery. The time constant is τ. What is the current in the circuit at time t = 1τ after the connection is made?

  • (A) 2.76 mA
  • (B) 5.52 mA
  • (C) 1.38 mA
  • (D) 4.14 mA
Show Answer

Answer: A – Solving the RC differential equation gives I(t) = (V/R)e-t/τ. Here V/R = 7.5 mA, so at t = 1τ, I = 7.5e-1 = 2.76 mA.

Q38. A capacitor charged to 12 V is discharged through a resistor. The time constant of the circuit is τ. What is the voltage across the capacitor at time t = 2τ after discharging begins?

  • (A) 3.25 V
  • (B) 1.62 V
  • (C) 0.812 V
  • (D) 2.44 V
Show Answer

Answer: B – From dV/dt = -V/τ, V(t) = V0e-t/τ = (12)e-2 = 1.62 V.

Q39. A wire of length 3 m and cross-sectional area 2 mm² is made of a material of resistivity 0.8×10-6 Ω·m. What is the resistance of the wire?

  • (A) 2.4 Ω
  • (B) 1.2 Ω
  • (C) 0.6 Ω
  • (D) 1.8 Ω
Show Answer

Answer: B – R = ρL/A = (0.8×10-6)(3)/(2×10-6) = 1.2 Ω.

Q40. A total current of 3 A splits between two parallel branches of resistance 2 Ω and 3 Ω. What is the current through the 2 Ω branch?

  • (A) 3.6 A
  • (B) 0.9 A
  • (C) 1.8 A
  • (D) 2.7 A
Show Answer

Answer: C – Both branches have the same voltage, so I1 = ItotalR2/(R1 + R2) = (3)(3)/(2 + 3) = 1.8 A.

Q41. A current of 5 A flows through a 5 Ω resistor for 120 s. How much electrical energy is converted to heat in the resistor during this time?

  • (A) 15 kJ
  • (B) 30 kJ
  • (C) 7.5 kJ
  • (D) 22.5 kJ
Show Answer

Answer: A – E = I²Rt = (5)²(5)(120) = 15 kJ.

Unit 12: Magnetic Fields and Electromagnetism

Q42. An electron moves at 2×106 m/s perpendicular to a uniform magnetic field of 0.01 T. What is the magnitude of the magnetic force on the electron?

  • (A) 6.4 ×10-15 N
  • (B) 1.6 ×10-15 N
  • (C) 4.8 ×10-15 N
  • (D) 3.2 ×10-15 N
Show Answer

Answer: D – F = qvB sinθ = (1.60×10-19)(2×106)(0.01) = 3.2×10-15 N.

Q43. A straight wire of length 1 m carries a current of 3 A perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the wire?

  • (A) 1.2 N
  • (B) 0.3 N
  • (C) 0.9 N
  • (D) 0.6 N
Show Answer

Answer: D – F = ILB sinθ = (3)(1)(0.2) = 0.6 N.

Q44. A long straight wire carries a current of 12 A. What is the magnitude of the magnetic field at a distance of 0.08 m from the wire?

  • (A) 30 μT
  • (B) 60 μT
  • (C) 15 μT
  • (D) 45 μT
Show Answer

Answer: A – B = μ0I/(2πr) = (2×10-7)(12)/(0.08) = 30 μT.

Q45. A circular loop of radius 0.05 m carries a current of 8 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 201 μT
  • (B) 50.3 μT
  • (C) 101 μT
  • (D) 151 μT
Show Answer

Answer: C – B = μ0I/(2R) = (4π×10-7)(8)/(2×0.05) = 101 μT.

Q46. A solenoid has 800 turns per metre and carries a current of 3 A. What is the magnitude of the magnetic field inside the solenoid, far from its ends?

  • (A) 6.03 mT
  • (B) 1.51 mT
  • (C) 4.52 mT
  • (D) 3.02 mT
Show Answer

Answer: D – B = μ0nI = (4π×10-7)(800)(3) = 3.02 mT.

Q47. A proton moves at 5×104 m/s perpendicular to a uniform magnetic field of 0.02 T. What is the radius of its circular path?

  • (A) 2.61 cm
  • (B) 5.22 cm
  • (C) 1.3 cm
  • (D) 3.91 cm
Show Answer

Answer: A – qvB = mv²/r, so r = mv/(qB) = (1.67×10-27)(5×104)/((1.60×10-19)(0.02)) = 2.61 cm.

Q48. A rectangular coil of 20 turns, of sides 0.1 m by 0.1 m, carries a current of 2 A in a uniform magnetic field of 0.3 T. When the angle between the coil’s normal and the field is 90°, what is the magnitude of the torque on the coil?

  • (A) 240 mN·m
  • (B) 60 mN·m
  • (C) 120 mN·m
  • (D) 180 mN·m
Show Answer

Answer: C – τ = NIAB sinθ = (20)(2)(0.1×0.1)(0.3) sin 90° = 120 mN·m.

Q49. Two long parallel wires carry currents of 6 A and 4 A in the same direction and are 0.2 m apart. What is the magnitude of the magnetic force per unit length between the wires?

  • (A) 0.024 mN/m
  • (B) 0.048 mN/m
  • (C) 0.012 mN/m
  • (D) 0.036 mN/m
Show Answer

Answer: A – F/L = μ0I1I2/(2πd) = (2×10-7)(6)(4)/(0.2) = 0.024 mN/m. The wires attract each other.

Q50. A long straight cylindrical wire of radius 0.05 m carries a uniformly distributed current of 8 A. What is the magnitude of the magnetic field at a distance of 0.02 m from the wire’s axis, inside the wire?

  • (A) 25.6 μT
  • (B) 6.4 μT
  • (C) 12.8 μT
  • (D) 19.2 μT
Show Answer

Answer: C – By Ampere’s law inside the wire, B = μ0Ir/(2πR²) = (2×10-7)(8)(0.02)/(0.05)² = 12.8 μT.

Unit 13: Electromagnetic Induction

Q51. A flat loop of area 100 cm² is placed in a uniform magnetic field of 0.15 T. The angle between the field and the normal to the loop is 60°. What is the magnetic flux through the loop?

  • (A) 1.5 mWb
  • (B) 0.75 mWb
  • (C) 0.375 mWb
  • (D) 1.13 mWb
Show Answer

Answer: B – Φ = BA cosθ = (0.15)(100×10-4) cos 60° = 0.75 mWb.

Q52. A coil of 20 turns and area 200 cm² is placed with its plane perpendicular to a magnetic field. The field strength increases uniformly at 0.5 T/s. What is the magnitude of the induced emf in the coil?

  • (A) 400 mV
  • (B) 100 mV
  • (C) 300 mV
  • (D) 200 mV
Show Answer

Answer: D – ε = -N dΦ/dt = NA(dB/dt) = (20)(200×10-4)(0.5) = 200 mV in magnitude.

Q53. A conducting rod of length 1 m moves at 10 m/s perpendicular to its length and perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the motional emf between the ends of the rod?

  • (A) 10 V
  • (B) 5 V
  • (C) 2.5 V
  • (D) 7.5 V
Show Answer

Answer: B – ε = BLv = (0.5)(1)(10) = 5 V.

Q54. A coil with 5 turns, area 400 cm² and resistance 40 Ω lies perpendicular to a magnetic field that changes at 2 T/s. What is the magnitude of the induced current in the coil?

  • (A) 10 mA
  • (B) 20 mA
  • (C) 5 mA
  • (D) 15 mA
Show Answer

Answer: A – ε = N A(dB/dt) = (5)(400×10-4)(2) V, so I = ε/R = 10 mA. By Lenz’s law its direction opposes the change in flux.

Q55. An inductor of inductance 50 mH carries a current that changes at a rate of 25 A/s. What is the magnitude of the induced emf across the inductor?

  • (A) 2.5 V
  • (B) 1.25 V
  • (C) 0.625 V
  • (D) 1.88 V
Show Answer

Answer: B – ε = L|dI/dt| = (50×10-3)(25) = 1.25 V, opposing the change in current.

Q56. An inductor of inductance 20 mH carries a steady current of 2 A. How much energy is stored in its magnetic field?

  • (A) 40 mJ
  • (B) 80 mJ
  • (C) 20 mJ
  • (D) 60 mJ
Show Answer

Answer: A – U = (1/2)LI² = (1/2)(20×10-3)(2)² = 40 mJ.

Q57. An inductor of inductance 100 mH is connected in series with a resistor of 20 Ω. What is the time constant of this LR circuit?

  • (A) 10 ms
  • (B) 5 ms
  • (C) 2.5 ms
  • (D) 7.5 ms
Show Answer

Answer: B – τ = L/R = (100×10-3)/(20) = 5 ms.

Q58. A series LR circuit with a 6 V battery and a resistor of 5 Ω is switched on at t = 0 with zero initial current. The time constant is τ. What is the current at t = 2τ?

  • (A) 2.08 A
  • (B) 0.519 A
  • (C) 1.04 A
  • (D) 1.56 A
Show Answer

Answer: C – Solving L(dI/dt) + RI = V gives I(t) = (V/R)(1 – e-t/τ). With V/R = 1.2 A, at t = 2τ, I = 1.04 A.

Q59. An LC circuit consists of an inductor of 10 mH and a capacitor of 5 μF. What is the frequency of oscillation of the circuit?

  • (A) 1.42 kHz
  • (B) 0.712 kHz
  • (C) 0.356 kHz
  • (D) 1.07 kHz
Show Answer

Answer: B – f = 1/(2π√(LC)) = 1/(2π√[(10×10-3)(5×10-6)]) = 0.712 kHz.

Q60. An LC circuit has inductance 5 mH and capacitance 4 μF. What is the angular frequency of its charge oscillations?

  • (A) 14.1 krad/s
  • (B) 3.54 krad/s
  • (C) 10.6 krad/s
  • (D) 7.07 krad/s
Show Answer

Answer: D – ω = 1/√(LC) = 1/√[(5×10-3)(4×10-6)] = 7.07 krad/s.

Well done for finishing the test. Review every question you missed, work through its explanation again, and revisit the unit where you lost the most marks before you try the next practice test.

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