AP Physics C: Electricity and Magnetism Practice Test 17 – 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 7 μC and 3 μC and are held 0.2 m apart in air. What is the magnitude of the electrostatic force between them?

  • (A) 9.45 N
  • (B) 2.36 N
  • (C) 7.09 N
  • (D) 4.72 N
Show Answer

Answer: D – Using Coulomb’s law, F = kq1q2/r² = (9.00×109)(7×10-6)(3×10-6)/(0.2)² = 4.72 N.

Q2. A point charge of 6 μC is fixed in vacuum. What is the magnitude of the electric field at a point 0.6 m from the charge?

  • (A) 150 kN/C
  • (B) 300 kN/C
  • (C) 75 kN/C
  • (D) 225 kN/C
Show Answer

Answer: A – E = kQ/r² = (9.00×109)(6×10-6)/(0.6)² = 1.5×105 N/C = 150 kN/C.

Q3. A closed Gaussian surface encloses a net charge of 6 nC. What is the net electric flux through the surface?

  • (A) 1.36×103 N·m²/C
  • (B) 678 N·m²/C
  • (C) 339 N·m²/C
  • (D) 1.02×103 N·m²/C
Show Answer

Answer: B – By Gauss’s law, Φ = Qenc/ε0 = (6×10-9)/(8.85×10-12) = 678 N·m²/C.

Q4. An infinitely long straight wire carries a uniform linear charge density of 1.5 μC/m. What is the magnitude of the electric field at a perpendicular distance of 0.3 m from the wire?

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

Answer: A – For an infinite line, E = 2kλ/r = 2(9.00×109)(1.5×10-6)/(0.3) = 90 kN/C.

Q5. A very large flat sheet carries a uniform surface charge density of 3 μC/m². What is the magnitude of the electric field just outside the sheet?

  • (A) 339 kN/C
  • (B) 169 kN/C
  • (C) 84.7 kN/C
  • (D) 254 kN/C
Show Answer

Answer: B – For an infinite sheet, E = σ/(2ε0) = (3×10-6)/(2×8.85×10-12) = 169 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 0.12 m from its centre?

  • (A) 2.16×103 kN/C
  • (B) 540 kN/C
  • (C) 1.08×103 kN/C
  • (D) 1.62×103 kN/C
Show Answer

Answer: C – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R³ = (9.00×109)(8×10-6)(0.12)/(0.2)³ = 1.08×103 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 1 m from the centre of the shell?

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

Answer: A – Outside the shell the field is that of a point charge at the centre: E = kQ/r² = (9.00×109)(2×10-6)/(1)² = 18 kN/C.

Q8. A thin rod of length 0.4 m carries a total charge of 8 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) 2.88×103 kN/C
  • (B) 720 kN/C
  • (C) 1.44×103 kN/C
  • (D) 2.16×103 kN/C
Show Answer

Answer: C – Integrating dE = kλdx/x² from x = 0.1 m to 0.5 m gives E = kQ/[a(a + L)] = (9.00×109)(8×10-6)/[(0.1)(0.5)] = 1.44×103 kN/C.

Q9. A thin ring of radius 0.25 m carries a total charge of 4 μC uniformly around its circumference. What is the magnitude of the electric field on the ring’s axis at a distance of 0.2 m from its centre?

  • (A) 439 kN/C
  • (B) 110 kN/C
  • (C) 219 kN/C
  • (D) 329 kN/C
Show Answer

Answer: C – On the axis, E = kQz/(z² + R²)3/2 = (9.00×109)(4×10-6)(0.2)/(0.2² + 0.25²)3/2 = 219 kN/C.

Q10. An electric dipole consists of charges of +4 μC and -4 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) 135 kN/C
  • (B) 33.7 kN/C
  • (C) 67.5 kN/C
  • (D) 101 kN/C
Show Answer

Answer: C – On the axis far from a dipole, E ≈ 2kp/r³ with p = qd = (4×10-6)(0.06) C·m, giving E = 67.5 kN/C.

Q11. A closed surface surrounds a dipole made of two equal and opposite charges, with no other charge enclosed. 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 2 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 0.35 m from the second charge. What is the magnitude of the net electric field at P?

  • (A) 247 kN/C
  • (B) 61.7 kN/C
  • (C) 185 kN/C
  • (D) 123 kN/C
Show Answer

Answer: D – Both fields point away from their charges, in the same direction at P, so E = kq1/r1² + kq2/r2² = 123 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 0.25 m from the charge, taking the potential at infinity to be zero?

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

Answer: A – V = kQ/r = (9.00×109)(5×10-6)/(0.25) = 180 kV.

Q14. Two point charges of 6 μC and 4 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) 2.16 J
  • (C) 0.54 J
  • (D) 1.62 J
Show Answer

Answer: A – U = kq1q2/r = (9.00×109)(6×10-6)(4×10-6)/(0.2) = 1.08 J.

Q15. A thin ring of radius 0.15 m carries a total charge of 3 μC uniformly around its circumference. What is the electric potential on the ring’s axis at a distance of 0.2 m from its centre?

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

Answer: B – Every element of charge is at the same distance √(z² + R²) from the point, so V = kQ/√(z² + R²) = 108 kV.

Q16. A thin rod of length 0.5 m carries a uniform linear charge density of 3 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) 96.8 kV
  • (B) 24.2 kV
  • (C) 48.4 kV
  • (D) 72.6 kV
Show Answer

Answer: C – V = ∪kλdx/x = kλ ln[(a + L)/a] = (9.00×109)(3×10-6) ln(0.6/0.1) = 48.4 kV.

Q17. A uniform electric field of 600 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) 60 V
  • (B) -120 V
  • (C) -60 V
  • (D) -30 V
Show Answer

Answer: C – ΔV = -EΔx = -(600)(0.1) = -60 V. The potential falls in the direction of the field.

Q18. The electric potential along the x-axis is V(x) = 60x² + 40x volts, where x is in metres. What is the x-component of the electric field at x = 1 m?

  • (A) 160 V/m
  • (B) -320 V/m
  • (C) -80 V/m
  • (D) -160 V/m
Show Answer

Answer: D – Ex = -dV/dx = -(120x + 40). At x = 1 m this gives Ex = -160 V/m.

Q19. A charge of 6 μC is moved through a potential increase of 2000 V by an external agent at constant speed. How much work does the external agent do?

  • (A) 24 mJ
  • (B) 12 mJ
  • (C) 6 mJ
  • (D) 18 mJ
Show Answer

Answer: B – W = qΔV = (6×10-6)(2000) = 12 mJ.

Q20. A conducting sphere of radius 0.2 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) 270 kV
  • (C) 67.5 kV
  • (D) 202 kV
Show Answer

Answer: A – Outside a charged sphere V = kQ/r, so at the surface V = kQ/R = (9.00×109)(3×10-6)/(0.2) = 135 kV.

Q21. At one point in space the electric potential happens to be zero. Which statement about the electric field at that same 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 200 cm² separated by 4 mm of air. What is its capacitance?

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

Answer: B – C = ε0A/d = (8.85×10-12)(200×10-4)/(4×10-3) = 44.2 pF.

Q23. An air-filled capacitor has capacitance 80 pF. It is completely filled with a dielectric of dielectric constant 3 while its geometry is unchanged. What is the new capacitance?

  • (A) 480 pF
  • (B) 120 pF
  • (C) 240 pF
  • (D) 360 pF
Show Answer

Answer: C – C = κC0 = (3)(80 pF) = 240 pF.

Q24. A capacitor of capacitance 5 μF is charged to 150 V. How much energy is stored in the capacitor?

  • (A) 0.0281 mJ
  • (B) 0.113 mJ
  • (C) 0.0563 mJ
  • (D) 0.0844 mJ
Show Answer

Answer: C – U = (1/2)CV2 = 0.5 × (5 × 10-6) × (150)2 = 0.0563 mJ.

Q25. Capacitors of 8 μF and 12 μF are connected in series. What is their equivalent capacitance?

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

Answer: A – 1/Ceq = 1/8 + 1/12, so Ceq = 8×12/(8 + 12) = 4.8 μF.

Q26. Capacitors of 5 μF, 4 μF and 1 μF are connected in parallel. What is their equivalent capacitance?

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

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

Q27. A capacitor of capacitance 8 μF is connected across a 25 V battery. What charge is stored on each plate?

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

Answer: A – Q = CV = (8 μF)(25 V) = 200 μC.

Q28. A parallel-plate capacitor has a potential difference of 300 V across plates separated by 3 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) 150 kV/m
  • (D) 100 kV/m
Show Answer

Answer: D – E = V/d = (300)/(3×10-3) = 100 kV/m.

Q29. A spherical capacitor consists of concentric conducting spheres of radii 0.06 m and 0.15 m. What is its capacitance?

  • (A) 11.1 pF
  • (B) 22.2 pF
  • (C) 5.56 pF
  • (D) 16.7 pF
Show Answer

Answer: A – C = 4πε0ab/(b – a) = (4π×8.85×10-12)(0.06)(0.15)/(0.15 – 0.06) = 11.1 pF.

Unit 11: Electric Circuits

Q30. A resistor of 2 Ω is connected across a 9 V battery of negligible internal resistance. What is the current through the resistor?

  • (A) 9 A
  • (B) 2.25 A
  • (C) 4.5 A
  • (D) 6.75 A
Show Answer

Answer: C – I = V/R = 9/2 = 4.5 A.


Q31. Resistors of 3 Ω, 4 Ω and 5 Ω are connected in series across a 18 V battery. What is the current in the circuit?

  • (A) 1.5 A
  • (B) 3 A
  • (C) 0.75 A
  • (D) 2.25 A
Show Answer

Answer: A – Req = 3 + 4 + 5 = 12 Ω, so I = V/Req = 18/12 = 1.5 A.

Q32. Resistors of 6 Ω and 12 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 2 Ω
  • (B) 8 Ω
  • (C) 4 Ω
  • (D) 6 Ω
Show Answer

Answer: C – 1/Req = 1/R1 + 1/R2, so Req = 4 Ω.

Q33. A 3 Ω resistor is connected across a 20 V supply. At what rate is electrical energy converted to heat in the resistor?

  • (A) 267 W
  • (B) 133 W
  • (C) 66.7 W
  • (D) 200 W
Show Answer

Answer: B – P = V²/R = (20)²/3 = 133 W.

Q34. In a single loop, two batteries of emf 15 V and 6 V oppose each other, and the total resistance in the loop is 3 Ω. What is the current in the loop?

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

Answer: B – Applying Kirchhoff’s loop rule, (15 – 6) = I(3), so I = 9/3 = 3 A, in the direction of the larger emf.

Q35. A battery of emf 18 V and internal resistance 2 Ω is connected to an external resistor of 8 Ω. What is the terminal voltage of the battery?

  • (A) 28.8 V
  • (B) 7.2 V
  • (C) 21.6 V
  • (D) 14.4 V
Show Answer

Answer: D – I = E/(R + r) = 18/(8 + 2) A, so V = E – Ir = 14.4 V.

Q36. A resistor of 20 kΩ and a capacitor of 10 μF are connected in series. What is the time constant of this RC circuit?

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

Answer: B – τ = RC = (20×103 Ω)(10×10-6 F) = 200 ms.

Q37. An uncharged capacitor is connected in series with a 3 kΩ resistor across a 6 V battery. The time constant is τ. What is the current in the circuit at time t = 1τ after the connection is made?

  • (A) 1.47 mA
  • (B) 0.736 mA
  • (C) 0.368 mA
  • (D) 1.1 mA
Show Answer

Answer: B – Solving the RC differential equation gives I(t) = (V/R)e-t/τ. Here V/R = 2 mA, so at t = 1τ, I = 2e-1 = 0.736 mA.

Q38. A capacitor charged to 20 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) 14.7 V
  • (B) 7.36 V
  • (C) 3.68 V
  • (D) 11 V
Show Answer

Answer: B – From dV/dt = -V/τ, V(t) = V0e-t/τ = (20)e-1 = 7.36 V.

Q39. A wire of length 4 m and cross-sectional area 0.5 mm² is made of a material of resistivity 0.8×10-6 Ω·m. What is the resistance of the wire?

  • (A) 12.8 Ω
  • (B) 3.2 Ω
  • (C) 9.6 Ω
  • (D) 6.4 Ω
Show Answer

Answer: D – R = ρL/A = (0.8×10-6)(4)/(0.5×10-6) = 6.4 Ω.

Q40. A total current of 2 A splits between two parallel branches of resistance 2 Ω and 12 Ω. What is the current through the 2 Ω branch?

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

Answer: C – Both branches have the same voltage, so I1 = ItotalR2/(R1 + R2) = (2)(12)/(2 + 12) = 1.71 A.

Q41. A current of 5 A flows through a 10 Ω resistor for 120 s. How much electrical energy is converted to heat in the resistor during this time?

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

Answer: C – E = I²Rt = (5)²(10)(120) = 30 kJ.

Unit 12: Magnetic Fields and Electromagnetism

Q42. An electron moves at 3×106 m/s perpendicular to a uniform magnetic field of 0.05 T. What is the magnitude of the magnetic force on the electron?

  • (A) 48 ×10-15 N
  • (B) 12 ×10-15 N
  • (C) 36 ×10-15 N
  • (D) 24 ×10-15 N
Show Answer

Answer: D – F = qvB sinθ = (1.60×10-19)(3×106)(0.05) = 24×10-15 N.

Q43. A straight wire of length 0.5 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) 2 N
  • (B) 1 N
  • (C) 0.5 N
  • (D) 1.5 N
Show Answer

Answer: B – F = ILB sinθ = (10)(0.5)(0.2) = 1 N.

Q44. A long straight wire carries a current of 8 A. What is the magnitude of the magnetic field at a distance of 0.1 m from the wire?

  • (A) 16 μT
  • (B) 32 μT
  • (C) 8 μT
  • (D) 24 μT
Show Answer

Answer: A – B = μ0I/(2πr) = (2×10-7)(8)/(0.1) = 16 μT.

Q45. A circular loop of radius 0.08 m carries a current of 6 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 23.6 μT
  • (B) 94.2 μT
  • (C) 47.1 μT
  • (D) 70.7 μT
Show Answer

Answer: C – At the centre of a loop, B = μ0I/(2R) = 47.1 μT.

Q46. A solenoid has 700 turns per metre and carries a current of 2 A. What is the magnetic field magnitude inside the solenoid, far from its ends?

  • (A) 0.88 mT
  • (B) 1.76 mT
  • (C) 3.52 mT
  • (D) 2.64 mT
Show Answer

Answer: B – B = μ0nI = (4π × 10-7)(700)(2) = 1.76 mT.

Q47. A proton moves at 10×104 m/s perpendicular to a uniform magnetic field of 0.04 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)(10×104)/((1.60×10-19)(0.04)) = 2.61 cm.

Q48. A rectangular coil of 50 turns, of sides 0.1 m by 0.05 m, carries a current of 3 A in a uniform magnetic field of 0.3 T. When the angle between the coil’s normal and the field is 30°, what is the magnitude of the torque on the coil?

  • (A) 112 mN·m
  • (B) 225 mN·m
  • (C) 56.2 mN·m
  • (D) 169 mN·m
Show Answer

Answer: A – τ = NIAB sinθ = (50)(3)(0.1×0.05)(0.3) sin 30° = 112 mN·m.

Q49. Two long parallel wires carry currents of 6 A and 6 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) 0.288 mN/m
  • (B) 0.072 mN/m
  • (C) 0.216 mN/m
  • (D) 0.144 mN/m
Show Answer

Answer: D – F/L = μ0I1I2/(2πd) = (2×10-7)(6)(6)/(0.05) = 0.144 mN/m. The wires attract each other.

Q50. A long straight cylindrical wire of radius 0.05 m carries a uniformly distributed current of 12 A. What is the magnitude of the magnetic field at a distance of 0.01 m from the wire’s axis, inside the wire?

  • (A) 19.2 μT
  • (B) 4.8 μT
  • (C) 9.6 μT
  • (D) 14.4 μT
Show Answer

Answer: C – By Ampere’s law inside the wire, B = μ0Ir/(2πR²) = (2×10-7)(12)(0.01)/(0.05)² = 9.6 μT.

Unit 13: Electromagnetic Induction

Q51. A flat loop of area 300 cm² is placed in a uniform magnetic field of 0.1 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 = (0.1)(300×10-4) cos 0° = 3 mWb.

Q52. A coil of 50 turns and area 200 cm² is placed with its plane perpendicular to a magnetic field. The field strength increases uniformly at 1 T/s. What is the magnitude of the induced emf in the coil?

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

Answer: D – ε = -N dΦ/dt = NA(dB/dt) = (50)(200×10-4)(1) = 1×103 mV in magnitude.

Q53. A conducting rod of length 0.4 m moves at 4 m/s perpendicular to its length and perpendicular to a uniform magnetic field of 0.4 T. What is the magnitude of the motional emf between the ends of the rod?

  • (A) 1.28 V
  • (B) 0.32 V
  • (C) 0.96 V
  • (D) 0.64 V
Show Answer

Answer: D – ε = BLv = (0.4)(0.4)(4) = 0.64 V.

Q54. A coil with 2 turns, area 200 cm² and resistance 10 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) 16 mA
  • (B) 8 mA
  • (C) 4 mA
  • (D) 12 mA
Show Answer

Answer: B – ε = N A(dB/dt) = (2)(200×10-4)(2) V, so I = ε/R = 8 mA. By Lenz’s law its direction opposes the change in flux.

Q55. An inductor of inductance 200 mH carries a current that changes at a rate of 50 A/s. What is the magnitude of the induced emf across the inductor?

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

Answer: D – ε = L|dI/dt| = (200×10-3)(50) = 10 V, opposing the change in current.

Q56. An inductor of inductance 200 mH carries a steady current of 5 A. How much energy is stored in its magnetic field?

  • (A) 5×103 mJ
  • (B) 2.5×103 mJ
  • (C) 1.25×103 mJ
  • (D) 3.75×103 mJ
Show Answer

Answer: B – U = (1/2)LI² = (1/2)(200×10-3)(5)² = 2.5×103 mJ.

Q57. An inductor of inductance 250 mH is connected in series with a resistor of 50 Ω. What is the time constant of this LR circuit?

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

Answer: D – τ = L/R = (250×10-3)/(50) = 5 ms.

Q58. A series LR circuit with a 6 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) 2.59 A
  • (B) 1.3 A
  • (C) 0.648 A
  • (D) 1.95 A
Show Answer

Answer: B – Solving L(dI/dt) + RI = V gives I(t) = (V/R)(1 – e-t/τ). With V/R = 1.5 A, at t = 2τ, I = 1.3 A.

Q59. An LC circuit consists of an inductor of 5 mH and a capacitor of 2 μF. What is the frequency of oscillation of the circuit?

  • (A) 3.18 kHz
  • (B) 0.796 kHz
  • (C) 1.59 kHz
  • (D) 2.39 kHz
Show Answer

Answer: C – f = 1/(2π√(LC)) = 1/(2π√[(5×10-3)(2×10-6)]) = 1.59 kHz.

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

  • (A) 15.8 krad/s
  • (B) 3.95 krad/s
  • (C) 11.9 krad/s
  • (D) 7.91 krad/s
Show Answer

Answer: D – ω = 1/√(LC) = 1/√[(4×10-3)(4×10-6)] = 7.91 krad/s.

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