AP Physics C: Electricity and Magnetism Practice Test 18 – 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 small charged spheres carry charges of 7 μC and 2 μC and are held 0.1 m apart in air. What is the magnitude of the electrostatic force between them?

  • (A) 25.2 N
  • (B) 6.3 N
  • (C) 18.9 N
  • (D) 12.6 N
Show Answer

Answer: D – Using Coulomb’s law, F = kq1q2/r² = (9.00×109)(7×10-6)(2×10-6)/(0.1)² = 12.6 N.

Q2. A point charge of 4 μC is fixed in vacuum. What is the magnitude of the electric field at a point 0.3 m from the charge?

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

Answer: C – E = kQ/r² = (9.00×109)(4×10-6)/(0.3)² = 4×105 N/C = 400 kN/C.

Q3. A closed Gaussian surface encloses a net charge of 9 nC. What is the net electric flux through the surface?

  • (A) 2.03×103 N·m²/C
  • (B) 508 N·m²/C
  • (C) 1.53×103 N·m²/C
  • (D) 1.02×103 N·m²/C
Show Answer

Answer: D – By Gauss’s law, Φ = Qenc/ε0 = (9×10-9)/(8.85×10-12) = 1.02×103 N·m²/C.

Q4. An infinitely long straight wire carries a uniform linear charge density of 1.2 μC/m. What is the magnitude of the electric field at a perpendicular distance of 0.4 m from the wire?

  • (A) 108 kN/C
  • (B) 27 kN/C
  • (C) 54 kN/C
  • (D) 81 kN/C
Show Answer

Answer: C – For an infinite line, E = 2kλ/r = 2(9.00×109)(1.2×10-6)/(0.4) = 54 kN/C.

Q5. A very large flat conducting sheet carries a uniform surface charge density of 5 μC/m². What is the magnitude of the electric field just outside the sheet?

  • (A) 282 kN/C
  • (B) 1.13×103 kN/C
  • (C) 565 kN/C
  • (D) 847 kN/C
Show Answer

Answer: C – Just outside a charged conducting sheet, E = σ/ε0 = 565 kN/C.

Q6. A solid insulating sphere of radius 0.2 m carries a total charge of 8 μC uniformly distributed through its volume. What is the magnitude of the electric field at a distance of 0.08 m from its centre?

  • (A) 720 kN/C
  • (B) 1.44×103 kN/C
  • (C) 360 kN/C
  • (D) 1.08×103 kN/C
Show Answer

Answer: A – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R³ = (9.00×109)(8×10-6)(0.08)/(0.2)³ = 720 kN/C.

Q7. A thin spherical shell of radius 0.25 m carries a total charge of 5 μC uniformly on its surface. What is the magnitude of the electric field at a distance of 0.5 m from the centre of the shell?

  • (A) 360 kN/C
  • (B) 180 kN/C
  • (C) 90 kN/C
  • (D) 270 kN/C
Show Answer

Answer: B – Outside the shell the field is that of a point charge at the centre: E = kQ/r² = (9.00×109)(5×10-6)/(0.5)² = 180 kN/C.

Q8. A thin rod of length 0.5 m carries a total charge of 6 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 0.25 m from its nearer end?

  • (A) 576 kN/C
  • (B) 288 kN/C
  • (C) 144 kN/C
  • (D) 432 kN/C
Show Answer

Answer: B – Integrating dE = kλdx/x² from x = 0.25 m to 0.75 m gives E = kQ/[a(a + L)] = (9.00×109)(6×10-6)/[(0.25)(0.75)] = 288 kN/C.

Q9. A thin ring of radius 0.2 m carries a total charge of 5 μC uniformly around its circumference. What is the magnitude of the electric field on the ring’s axis at a distance of 0.4 m from its centre?

  • (A) 402 kN/C
  • (B) 201 kN/C
  • (C) 101 kN/C
  • (D) 302 kN/C
Show Answer

Answer: B – On the axis, E = kQz/(z² + R²)3/2 = (9.00×109)(5×10-6)(0.4)/(0.4² + 0.2²)3/2 = 201 kN/C.

Q10. An electric dipole consists of charges of +5 μC and -5 μC separated by 0.06 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) 169 kN/C
  • (B) 42.2 kN/C
  • (C) 127 kN/C
  • (D) 84.4 kN/C
Show Answer

Answer: D – On the axis far from a dipole, E ≈ 2kp/r³ with p = qd = (5×10-6)(0.06) C·m, giving E = 84.4 kN/C.

Q11. A Gaussian surface encloses a pair of charges +q and -q and no other charge. 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 enclosed charges +q and -q cancel, 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 2 μC and 3 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 0.35 m from the second charge. What is the magnitude of the net electric field at P?

  • (A) 541 kN/C
  • (B) 270 kN/C
  • (C) 135 kN/C
  • (D) 406 kN/C
Show Answer

Answer: B – Both fields point away from their charges, in the same direction at P, so E = kq1/r1² + kq2/r2² = 270 kN/C.

Unit 9: Electric Potential

Q13. A point charge of 3 μC is fixed in vacuum. What is the electric potential at a distance of 0.25 m from the charge, taking the potential at infinity to be zero?

  • (A) 108 kV
  • (B) 216 kV
  • (C) 54 kV
  • (D) 162 kV
Show Answer

Answer: A – V = kQ/r = (9.00×109)(3×10-6)/(0.25) = 108 kV.

Q14. Two point charges of 3 μC and 2 μC are held 0.2 m apart. What is the electric potential energy of this pair, taking zero energy at infinite separation?

  • (A) 0.54 J
  • (B) 0.135 J
  • (C) 0.405 J
  • (D) 0.27 J
Show Answer

Answer: D – U = kq1q2/r = (9.00×109)(3×10-6)(2×10-6)/(0.2) = 0.27 J.

Q15. A thin ring of radius 0.1 m carries a total charge of 2 μC uniformly around its circumference. What is the electric potential on the ring’s axis at a distance of 0.3 m from its centre?

  • (A) 114 kV
  • (B) 28.5 kV
  • (C) 85.4 kV
  • (D) 56.9 kV
Show Answer

Answer: D – Every element of charge is at the same distance √(z² + R²) from the point, so V = kQ/√(z² + R²) = 56.9 kV.

Q16. A thin rod of length 0.5 m carries a uniform linear charge density of 2 μC/m. By integrating the contributions of its charge elements, what is the electric potential at a point on the rod’s axis 0.1 m from its nearer end, with zero potential at infinity?

  • (A) 64.5 kV
  • (B) 16.1 kV
  • (C) 32.3 kV
  • (D) 48.4 kV
Show Answer

Answer: C – V = ∪kλdx/x = kλ ln[(a + L)/a] = (9.00×109)(2×10-6) ln(0.6/0.1) = 32.3 kV.

Q17. A uniform electric field of 800 N/C points in the +x direction. A charge is moved 0.15 m in the +x direction. By how much does the electric potential change?

  • (A) -120 V
  • (B) 120 V
  • (C) -240 V
  • (D) -60 V
Show Answer

Answer: A – ΔV = -EΔx = -(800)(0.15) = -120 V. The potential falls in the direction of the field.

Q18. The electric potential along the x-axis is V(x) = 60x² + 30x volts, where x is in metres. What is the x-component of the electric field at x = 2 m?

  • (A) 270 V/m
  • (B) -540 V/m
  • (C) -135 V/m
  • (D) -270 V/m
Show Answer

Answer: D – Ex = -dV/dx = -(120x + 30). At x = 2 m this gives Ex = -270 V/m.

Q19. A charge of 6 μC is moved through a potential increase of 4000 V by an external agent at constant speed. How much work does the external agent do?

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

Answer: A – W = qΔV = (6×10-6)(4000) = 24 mJ.

Q20. A conducting sphere of radius 0.4 m carries a total charge of 3 μC. What is the electric potential at the surface of the sphere, with zero potential at infinity?

  • (A) 135 kV
  • (B) 33.8 kV
  • (C) 67.5 kV
  • (D) 101 kV
Show Answer

Answer: C – Outside a charged sphere V = kQ/r, so at the surface V = kQ/R = (9.00×109)(3×10-6)/(0.4) = 67.5 kV.

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

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

Answer: C – 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 location 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 400 cm² separated by 3 mm of air. What is its capacitance?

  • (A) 236 pF
  • (B) 59 pF
  • (C) 177 pF
  • (D) 118 pF
Show Answer

Answer: D – C = ε0A/d = (8.85×10-12)(400×10-4)/(3×10-3) = 118 pF.

Q23. An air-filled capacitor has capacitance 20 pF. It is completely filled with a dielectric of dielectric constant 3 while its geometry is unchanged. What is the new capacitance?

  • (A) 120 pF
  • (B) 30 pF
  • (C) 60 pF
  • (D) 90 pF
Show Answer

Answer: C – C = κC0 = (3)(20 pF) = 60 pF.

Q24. A capacitor of capacitance 4 μF is charged to 50 V. How much energy is stored in the capacitor?

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

Answer: C – U = (1/2)CV² = (1/2)(4×10-6)(50)² = 5 mJ.

Q25. Capacitors of 4 μF and 2 μF are connected in series. What is their equivalent capacitance?

  • (A) 2.67 μF
  • (B) 1.33 μF
  • (C) 0.667 μF
  • (D) 2 μF
Show Answer

Answer: B – 1/Ceq = 1/4 + 1/2, so Ceq = 4×2/(4 + 2) = 1.33 μF.

Q26. Capacitors of 2 μF, 3 μF and 1 μF are connected in parallel. What is their equivalent capacitance?

  • (A) 6 μF
  • (B) 12 μF
  • (C) 3 μF
  • (D) 9 μF
Show Answer

Answer: A – For a parallel combination, Ceq = C1 + C2 + C3 = 6 μF.

Q27. A capacitor of capacitance 6 μF is connected across a 100 V battery. What charge is stored on each plate?

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

Answer: A – Q = CV = (6 μF)(100 V) = 600 μC.

Q28. A parallel-plate capacitor has a potential difference of 400 V across plates separated by 4 mm. What is the magnitude of the uniform electric field between the plates, ignoring edge effects?

  • (A) 200 kV/m
  • (B) 50 kV/m
  • (C) 100 kV/m
  • (D) 150 kV/m
Show Answer

Answer: C – E = V/d = (400)/(4×10-3) = 100 kV/m.

Q29. A spherical capacitor consists of concentric conducting spheres of radii 0.06 m and 0.2 m. What is its capacitance?

  • (A) 19.1 pF
  • (B) 4.77 pF
  • (C) 14.3 pF
  • (D) 9.53 pF
Show Answer

Answer: D – C = 4πε0ab/(b – a) = (4π×8.85×10-12)(0.06)(0.2)/(0.2 – 0.06) = 9.53 pF.

Unit 11: Electric Circuits

Q30. A resistor of 6 Ω is connected across a 12 V battery of negligible internal resistance. What is the current through the resistor?

  • (A) 4 A
  • (B) 2 A
  • (C) 1 A
  • (D) 3 A
Show Answer

Answer: B – I = V/R = 12/6 = 2 A.


Q31. Resistors of 3 Ω, 4 Ω and 4 Ω are connected in series across a 24 V battery. What is the current in the circuit?

  • (A) 2.18 A
  • (B) 4.36 A
  • (C) 1.09 A
  • (D) 3.27 A
Show Answer

Answer: A – Req = 3 + 4 + 4 = 11 Ω, so I = V/Req = 24/11 = 2.18 A.

Q32. Resistors of 12 Ω and 8 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 9.6 Ω
  • (B) 4.8 Ω
  • (C) 2.4 Ω
  • (D) 7.2 Ω
Show Answer

Answer: B – 1/Req = 1/12 + 1/8, so Req = 12×8/(12 + 8) = 4.8 Ω.

Q33. A 4 Ω resistor is connected across a 20 V supply. At what rate is electrical energy converted to heat in the resistor?

  • (A) 200 W
  • (B) 50 W
  • (C) 100 W
  • (D) 150 W
Show Answer

Answer: C – P = V²/R = (20)²/4 = 100 W.

Q34. In a single loop, two batteries of emf 12 V and 3 V oppose each other, and the total resistance in the loop is 4 Ω. What is the current in the loop?

  • (A) 2.25 A
  • (B) 4.5 A
  • (C) 1.12 A
  • (D) 3.38 A
Show Answer

Answer: A – Applying Kirchhoff’s loop rule, (12 – 3) = I(4), so I = 9/4 = 2.25 A, in the direction of the larger emf.

Q35. A battery of emf 9 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. What is the terminal voltage of the battery?

  • (A) 7.5 V
  • (B) 15 V
  • (C) 3.75 V
  • (D) 11.2 V
Show Answer

Answer: A – I = E/(R + r) = 9/(5 + 1) A, so V = E – Ir = 7.5 V.

Q36. A resistor of 25 kΩ and a capacitor of 5 μF are connected in series. What is the time constant of this RC circuit?

  • (A) 250 ms
  • (B) 125 ms
  • (C) 62.5 ms
  • (D) 188 ms
Show Answer

Answer: B – τ = RC = (25×103 Ω)(5×10-6 F) = 125 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 = 2τ after the connection is made?

  • (A) 2.03 mA
  • (B) 1.02 mA
  • (C) 0.508 mA
  • (D) 1.52 mA
Show Answer

Answer: B – Solving the RC differential equation gives I(t) = (V/R)e-t/τ. Here V/R = 7.5 mA, so at t = 2τ, I = 7.5e-2 = 1.02 mA.

Q38. A capacitor charged to 25 V is discharged through a resistor. The time constant of the circuit is τ. What is the voltage across the capacitor at time t = 1τ after discharging begins?

  • (A) 4.6 V
  • (B) 18.4 V
  • (C) 13.8 V
  • (D) 9.2 V
Show Answer

Answer: D – From dV/dt = -V/τ, V(t) = V0e-t/τ = (25)e-1 = 9.2 V.

Q39. A wire of length 2 m and cross-sectional area 0.5 mm² is made of a material of resistivity 0.5×10-6 Ω·m. What is the resistance of the wire?

  • (A) 4 Ω
  • (B) 1 Ω
  • (C) 3 Ω
  • (D) 2 Ω
Show Answer

Answer: D – R = ρL/A = (0.5×10-6)(2)/(0.5×10-6) = 2 Ω.

Q40. A total current of 4 A splits between two parallel branches of resistance 4 Ω and 3 Ω. What is the current through the 4 Ω branch?

  • (A) 1.71 A
  • (B) 3.43 A
  • (C) 0.857 A
  • (D) 2.57 A
Show Answer

Answer: A – Both branches have the same voltage, so I1 = ItotalR2/(R1 + R2) = (4)(3)/(4 + 3) = 1.71 A.

Q41. A current of 2 A flows through a 10 Ω resistor for 30 s. How much electrical energy is converted to heat in the resistor during this time?

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

Answer: B – E = I²Rt = (2)²(10)(30) = 1.2 kJ.

Unit 12: Magnetic Fields and Electromagnetism

Q42. An electron moves at 1×106 m/s perpendicular to a uniform magnetic field of 0.02 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)(1×106)(0.02) = 3.2×10-15 N.

Q43. A straight wire of length 1 m carries a current of 10 A perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the wire?

  • (A) 4 N
  • (B) 1 N
  • (C) 2 N
  • (D) 3 N
Show Answer

Answer: C – F = ILB sinθ = (10)(1)(0.2) = 2 N.

Q44. A long straight wire carries a current of 12 A. What is the magnitude of the magnetic field at a distance of 0.1 m from the wire?

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

Answer: B – B = μ0I/(2πr) = (2×10-7)(12)/(0.1) = 24 μT.

Q45. A circular loop of radius 0.04 m carries a current of 5 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 157 μT
  • (B) 39.3 μT
  • (C) 118 μT
  • (D) 78.5 μT
Show Answer

Answer: D – B = μ0I/(2R) = (4π×10-7)(5)/(2×0.04) = 78.5 μT.

Q46. A solenoid has 1000 turns per metre and carries a current of 4 A. What is the magnitude of the magnetic field inside the solenoid, far from its ends?

  • (A) 10.1 mT
  • (B) 2.51 mT
  • (C) 5.03 mT
  • (D) 7.54 mT
Show Answer

Answer: C – B = μ0nI = (4π×10-7)(1000)(4) = 5.03 mT.

Q47. A proton moves at 5×104 m/s perpendicular to a uniform magnetic field of 0.05 T. What is the radius of its circular path?

  • (A) 2.09 cm
  • (B) 0.522 cm
  • (C) 1.04 cm
  • (D) 1.57 cm
Show Answer

Answer: C – qvB = mv²/r, so r = mv/(qB) = (1.67×10-27)(5×104)/((1.60×10-19)(0.05)) = 1.04 cm.

Q48. A rectangular coil of 100 turns, of sides 0.1 m by 0.1 m, carries a current of 1 A in a uniform magnetic field of 0.3 T. When the angle between the coil’s normal and the field is 60°, what is the magnitude of the torque on the coil?

  • (A) 520 mN·m
  • (B) 130 mN·m
  • (C) 260 mN·m
  • (D) 390 mN·m
Show Answer

Answer: C – τ = NIAB sinθ = (100)(1)(0.1×0.1)(0.3) sin 60° = 260 mN·m.

Q49. Two long parallel wires carry currents of 8 A and 10 A in the same direction and are 0.05 m apart. What is the magnitude of the magnetic force per unit length between the wires?

  • (A) 0.64 mN/m
  • (B) 0.32 mN/m
  • (C) 0.16 mN/m
  • (D) 0.48 mN/m
Show Answer

Answer: B – F/L = μ0I1I2/(2πd) = (2×10-7)(8)(10)/(0.05) = 0.32 mN/m. The wires attract each other.

Q50. A long straight cylindrical wire of radius 0.06 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) 17.8 μT
  • (B) 4.44 μT
  • (C) 13.3 μT
  • (D) 8.89 μT
Show Answer

Answer: D – By Ampere’s law inside the wire, B = μ0Ir/(2πR²) = (2×10-7)(8)(0.02)/(0.06)² = 8.89 μT.

Unit 13: Electromagnetic Induction

Q51. A flat loop of area 100 cm² is placed in a uniform magnetic field of 0.3 T. The angle between the field and the normal to the loop is 0°. What is the magnetic flux through the loop?

  • (A) 6 mWb
  • (B) 1.5 mWb
  • (C) 4.5 mWb
  • (D) 3 mWb
Show Answer

Answer: D – Φ = BA cosθ = (0.3)(100×10-4) cos 0° = 3 mWb.

Q52. A coil of 25 turns and area 200 cm² is placed with its plane perpendicular to a magnetic field. The field strength increases uniformly at 2 T/s. What is the magnitude of the induced emf in the coil?

  • (A) 2×103 mV
  • (B) 1×103 mV
  • (C) 500 mV
  • (D) 1.5×103 mV
Show Answer

Answer: B – ε = -N dΦ/dt = NA(dB/dt) = (25)(200×10-4)(2) = 1×103 mV in magnitude.

Q53. A conducting rod of length 0.2 m moves at 2 m/s perpendicular to its length and perpendicular to a uniform magnetic field of 0.1 T. What is the magnitude of the motional emf between the ends of the rod?

  • (A) 0.08 V
  • (B) 0.02 V
  • (C) 0.06 V
  • (D) 0.04 V
Show Answer

Answer: D – ε = BLv = (0.1)(0.2)(2) = 0.04 V.

Q54. A coil with 1 turn, area 200 cm² and resistance 20 Ω lies perpendicular to a magnetic field that changes at 1 T/s. What is the magnitude of the induced current in the coil?

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

Answer: A – ε = N A(dB/dt) = (1)(200×10-4)(1) V, so I = ε/R = 1 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 20 A/s. What is the magnitude of the induced emf across the inductor?

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

Answer: B – ε = L|dI/dt| = (50×10-3)(20) = 1 V, opposing the change in current.

Q56. An inductor of inductance 50 mH carries a steady current of 3 A. How much energy is stored in its magnetic field?

  • (A) 450 mJ
  • (B) 225 mJ
  • (C) 112 mJ
  • (D) 338 mJ
Show Answer

Answer: B – U = (1/2)LI² = (1/2)(50×10-3)(3)² = 225 mJ.

Q57. An inductor of inductance 150 mH is connected in series with a resistor of 25 Ω. What is the time constant of this LR circuit?

  • (A) 3 ms
  • (B) 12 ms
  • (C) 6 ms
  • (D) 9 ms
Show Answer

Answer: C – τ = L/R = (150×10-3)/(25) = 6 ms.

Q58. A series LR circuit with a 20 V battery and a resistor of 4 Ω is switched on at t = 0 with zero initial current. The time constant is τ. What is the current at t = 2τ?

  • (A) 4.32 A
  • (B) 8.65 A
  • (C) 2.16 A
  • (D) 6.48 A
Show Answer

Answer: A – Solving L(dI/dt) + RI = V gives I(t) = (V/R)(1 – e-t/τ). With V/R = 5 A, at t = 2τ, I = 4.32 A.

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

  • (A) 1.01 kHz
  • (B) 2.01 kHz
  • (C) 0.503 kHz
  • (D) 1.51 kHz
Show Answer

Answer: A – f = 1/(2π√(LC)) = 1/(2π√[(5×10-3)(5×10-6)]) = 1.01 kHz.

Q60. An LC circuit has inductance 8 mH and capacitance 5 μF. What is the angular frequency of its charge oscillations?

  • (A) 5 krad/s
  • (B) 2.5 krad/s
  • (C) 10 krad/s
  • (D) 7.5 krad/s
Show Answer

Answer: A – ω = 1/√(LC) = 1/√[(8×10-3)(5×10-6)] = 5 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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Download Printable PDF – Test 18

Free printable version: all 60 questions + answer key. Explanations stay here on the website.