AP Physics C: Electricity and Magnetism Practice Test 19 – 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 11 μC and are held 0.3 m apart in air. What is the magnitude of the electrostatic force between them?

  • (A) 15.4 N
  • (B) 3.85 N
  • (C) 7.7 N
  • (D) 11.6 N
Show Answer

Answer: C – Using Coulomb’s law, F = kq1q2/r² = (9.00×109)(7×10-6)(11×10-6)/(0.3)² = 7.7 N.

Q2. A point charge of 2 μC is fixed in vacuum. What is the magnitude of the electric field at a point 0.3 m from the charge?

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

Answer: D – E = kQ/r² = (9.00×109)(2×10-6)/(0.3)² = 2×105 N/C = 200 kN/C.

Q3. A closed Gaussian surface encloses a net charge of 7 nC. What is the net electric flux through the surface?

  • (A) 791 N·m²/C
  • (B) 1.58×103 N·m²/C
  • (C) 395 N·m²/C
  • (D) 1.19×103 N·m²/C
Show Answer

Answer: A – By Gauss’s law, Φ = Qenc/ε0 = (7×10-9)/(8.85×10-12) = 791 N·m²/C.

Q4. An infinitely long straight wire carries a uniform linear charge density of 1 μC/m. What is the magnitude of the electric field at a perpendicular distance of 0.3 m from the wire?

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

Answer: A – For an infinite line, E = 2kλ/r = 2(9.00×109)(1×10-6)/(0.3) = 60 kN/C.

Q5. A very large flat sheet carries a uniform surface charge density of 2 μC/m². What is the magnitude of the electric field just outside the sheet?

  • (A) 226 kN/C
  • (B) 56.5 kN/C
  • (C) 169 kN/C
  • (D) 113 kN/C
Show Answer

Answer: D – For an infinite sheet, E = σ/(2ε0) = (2×10-6)/(2×8.85×10-12) = 113 kN/C.

Q6. A solid insulating sphere of radius 0.3 m carries a total charge of 4 μC uniformly distributed through its volume. What is the magnitude of the electric field at a distance of 0.05 m from its centre?

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

Answer: D – Inside a uniformly charged solid sphere, Gauss’s law gives E = kQr/R³ = (9.00×109)(4×10-6)(0.05)/(0.3)³ = 66.7 kN/C.

Q7. A thin spherical shell of radius 0.25 m carries a total charge of 4 μC uniformly on its surface. What is the magnitude of the electric field at a distance of 0.4 m from the centre of the shell?

  • (A) 450 kN/C
  • (B) 225 kN/C
  • (C) 112 kN/C
  • (D) 337 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)(4×10-6)/(0.4)² = 225 kN/C.

Q8. A thin rod of length 1 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) 346 kN/C
  • (B) 86.4 kN/C
  • (C) 173 kN/C
  • (D) 259 kN/C
Show Answer

Answer: C – Integrating dE = kλdx/x² from x = 0.25 m to 1.25 m gives E = kQ/[a(a + L)] = (9.00×109)(6×10-6)/[(0.25)(1.25)] = 173 kN/C.

Q9. A thin ring of radius 0.1 m carries a total charge of 6 μC uniformly around its circumference. What is the magnitude of the electric field on the ring’s axis at a distance of 0.3 m from its centre?

  • (A) 1.02×103 kN/C
  • (B) 256 kN/C
  • (C) 512 kN/C
  • (D) 768 kN/C
Show Answer

Answer: C – On the axis, E = kQz/(z² + R²)3/2 = (9.00×109)(6×10-6)(0.3)/(0.3² + 0.1²)3/2 = 512 kN/C.

Q10. An electric dipole consists of charges of +6 μC and -6 μC separated by 0.08 m. What is the approximate magnitude of the electric field on the dipole axis at a distance of 0.5 m from its centre, where the distance is much larger than the separation?

  • (A) 69 kN/C
  • (B) 34.5 kN/C
  • (C) 138 kN/C
  • (D) 104 kN/C
Show Answer

Answer: A – On the axis far from a dipole, E ≈ 2kp/r³ with p = qd = (6×10-6)(0.08) C·m, giving E = 69 kN/C.

Q11. A closed surface contains only an electric dipole, with equal positive and negative charges inside and nothing else. What is the net electric flux through the surface, and why?

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

Answer: D – Gauss’s law gives Φ = Qenc/ε0. Only the dipole is inside, so its +q and -q cancel to 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 0.25 m from the second charge. What is the magnitude of the net electric field at P?

  • (A) 964 kN/C
  • (B) 241 kN/C
  • (C) 723 kN/C
  • (D) 482 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 = 482 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 0.4 m from the charge, taking the potential at infinity to be zero?

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

Answer: A – V = kQ/r = (9.00×109)(3×10-6)/(0.4) = 67.5 kV.

Q14. Two point charges of 2 μC and 3 μC are held 0.5 m apart. What is the electric potential energy of this pair, taking zero energy at infinite separation?

  • (A) 0.216 J
  • (B) 0.108 J
  • (C) 0.054 J
  • (D) 0.162 J
Show Answer

Answer: B – U = kq1q2/r = (9.00×109)(2×10-6)(3×10-6)/(0.5) = 0.108 J.

Q15. A thin ring of radius 0.15 m carries a total charge of 6 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) 253 kV
  • (B) 63.2 kV
  • (C) 190 kV
  • (D) 126 kV
Show Answer

Answer: D – Every element of charge is at the same distance √(z² + R²) from the point, so V = kQ/√(z² + R²) = 126 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 0.1 m from its nearer end, with zero potential at infinity?

  • (A) 32.3 kV
  • (B) 8.06 kV
  • (C) 24.2 kV
  • (D) 16.1 kV
Show Answer

Answer: D – V = ∪kλdx/x = kλ ln[(a + L)/a] = (9.00×109)(1×10-6) ln(0.6/0.1) = 16.1 kV.

Q17. A uniform electric field of 600 N/C points in the +x direction. A charge is moved 0.25 m in the +x direction. By how much does the electric potential change?

  • (A) -150 V
  • (B) -75 V
  • (C) -300 V
  • (D) -225 V
Show Answer

Answer: A – ΔV = -EΔx = -(600)(0.25) = -150 V. The potential falls in the direction of the field.

Q18. The electric potential along the x-axis is V(x) = 50x² + 30x volts, where x is in metres. What is the x-component of the electric field at x = 1 m?

  • (A) -130 V/m
  • (B) 130 V/m
  • (C) -260 V/m
  • (D) -65 V/m
Show Answer

Answer: A – Ex = -dV/dx = -(100x + 30). At x = 1 m this gives Ex = -130 V/m.

Q19. A charge of 6 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) 60 mJ
  • (B) 30 mJ
  • (C) 15 mJ
  • (D) 45 mJ
Show Answer

Answer: B – W = qΔV = (6×10-6)(5000) = 30 mJ.

Q20. A conducting sphere of radius 0.25 m carries a total charge of 2 μC. What is the electric potential at the surface of the sphere, with zero potential at infinity?

  • (A) 144 kV
  • (B) 36 kV
  • (C) 108 kV
  • (D) 72 kV
Show Answer

Answer: D – Outside a charged sphere V = kQ/r, so at the surface V = kQ/R = (9.00×109)(2×10-6)/(0.25) = 72 kV.

Q21. At a point where the electric potential equals zero, which statement about the electric field there is necessarily correct?

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

Answer: B – Potential is defined up to an additive constant and depends on the reference point, while the field is the negative gradient of the potential. Zero potential at one point 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 2.5 mm of air. What is its capacitance?

  • (A) 35.4 pF
  • (B) 70.8 pF
  • (C) 142 pF
  • (D) 106 pF
Show Answer

Answer: B – C = ε0A/d = (8.85×10-12)(200×10-4)/(2.5×10-3) = 70.8 pF.

Q23. An air-filled capacitor has capacitance 20 pF. It is completely filled with a dielectric of dielectric constant 2 while its geometry is unchanged. What is the new capacitance?

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

Answer: D – C = κC0 = (2)(20 pF) = 40 pF.

Q24. A capacitor of capacitance 10 μF is charged to 150 V. How much energy is stored in the capacitor?

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

Answer: C – U = (1/2)CV² = (1/2)(10×10-6)(150)² = 112 mJ.

Q25. Capacitors of 10 μF and 5 μF are connected in series. What is their equivalent capacitance?

  • (A) 1.67×106 μF
  • (B) 3.33×106 μF
  • (C) 6.67×106 μF
  • (D) 5×106 μF
Show Answer

Answer: B – 1/Ceq = 1/C1 + 1/C2, so Ceq = 3.33×106 μF.

Q26. Capacitors of 2 μF, 7 μF and 1 μF are connected in parallel. What is their equivalent capacitance?

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

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

Q27. A capacitor of capacitance 4 μF is connected across a 50 V battery. What charge is stored on each plate?

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

Answer: D – Q = CV = (4 μF)(50 V) = 200 μC.

Q28. A parallel-plate capacitor has a potential difference of 100 V across plates separated by 3 mm. What is the magnitude of the uniform electric field between the plates, ignoring edge effects?

  • (A) 33.3 kV/m
  • (B) 66.7 kV/m
  • (C) 16.7 kV/m
  • (D) 50 kV/m
Show Answer

Answer: A – E = V/d = (100)/(3×10-3) = 33.3 kV/m.

Q29. A spherical capacitor consists of concentric conducting spheres of radii 0.06 m and 0.1 m. What is its capacitance?

  • (A) 16.7 pF
  • (B) 33.4 pF
  • (C) 8.34 pF
  • (D) 25 pF
Show Answer

Answer: A – C = 4πε0ab/(b – a) = (4π×8.85×10-12)(0.06)(0.1)/(0.1 – 0.06) = 16.7 pF.

Unit 11: Electric Circuits

Q30. A resistor of 2 Ω is connected across a 20 V battery of negligible internal resistance. What is the current through the resistor?

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

Answer: B – I = V/R = 20/2 = 10 A.


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

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

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

Q32. Resistors of 24 Ω and 8 Ω are connected in parallel. What is their equivalent resistance?

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

Answer: A – 1/Req = 1/R1 + 1/R2, so Req = 6 Ω.

Q33. A 5 Ω resistor is connected across a 12 V supply. At what rate is electrical energy converted to heat in the resistor?

  • (A) 28.8 W
  • (B) 57.6 W
  • (C) 14.4 W
  • (D) 43.2 W
Show Answer

Answer: A – P = V²/R = (12)²/5 = 28.8 W.

Q34. In a single loop, two batteries of emf 15 V and 4 V oppose each other, and the total resistance in the loop is 5 Ω. What is the current in the loop?

  • (A) 2.2 A
  • (B) 1.1 A
  • (C) 4.4 A
  • (D) 3.3 A
Show Answer

Answer: A – Applying Kirchhoff’s loop rule, (15 – 4) = I(5), so I = 11/5 = 2.2 A, in the direction of the larger emf.

Q35. A battery of emf 15 V and internal resistance 1 Ω is connected to an external resistor of 17 Ω. What is the terminal voltage of the battery?

  • (A) 28.3 V
  • (B) 7.08 V
  • (C) 21.2 V
  • (D) 14.2 V
Show Answer

Answer: D – I = E/(R + r) = 15/(17 + 1) A, so V = E – Ir = 14.2 V.

Q36. A resistor of 25 kΩ and a capacitor of 2 μF are connected in series. What is the time constant of this RC circuit?

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

Answer: B – τ = RC = (25×103 Ω)(2×10-6 F) = 50 ms.

Q37. An uncharged capacitor is connected in series with a 3 kΩ resistor across a 9 V battery. The time constant is τ. What is the current in the circuit at time t = 1τ after the connection is made?

  • (A) 1.1 mA
  • (B) 2.21 mA
  • (C) 0.552 mA
  • (D) 1.66 mA
Show Answer

Answer: A – Solving the RC differential equation gives I(t) = (V/R)e-t/τ. Here V/R = 3 mA, so at t = 1τ, I = 3e-1 = 1.1 mA.

Q38. A capacitor charged to 16 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) 2.94 V
  • (B) 11.8 V
  • (C) 8.83 V
  • (D) 5.89 V
Show Answer

Answer: D – From dV/dt = -V/τ, V(t) = V0e-t/τ = (16)e-1 = 5.89 V.

Q39. A wire of length 4 m and cross-sectional area 1 mm² is made of a material of resistivity 1.1×10-6 Ω·m. What is the resistance of the wire?

  • (A) 4.4 Ω
  • (B) 8.8 Ω
  • (C) 2.2 Ω
  • (D) 6.6 Ω
Show Answer

Answer: A – R = ρL/A = (1.1×10-6)(4)/(1×10-6) = 4.4 Ω.

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

  • (A) 4.57 A
  • (B) 2.29 A
  • (C) 1.14 A
  • (D) 3.43 A
Show Answer

Answer: B – Both branches have the same voltage, so I1 = ItotalR2/(R1 + R2) = (4)(4)/(3 + 4) = 2.29 A.

Q41. A current of 3 A flows through a 20 Ω resistor for 60 s. How much electrical energy is converted to heat in the resistor during this time?

  • (A) 10.8 kJ
  • (B) 21.6 kJ
  • (C) 5.4 kJ
  • (D) 16.2 kJ
Show Answer

Answer: A – E = I²Rt = (3)²(20)(60) = 10.8 kJ.

Unit 12: Magnetic Fields and Electromagnetism

Q42. An electron moves at 2×106 m/s perpendicular to a uniform magnetic field of 0.04 T. What is the magnitude of the magnetic force on the electron?

  • (A) 12.8 ×10-15 N
  • (B) 6.4 ×10-15 N
  • (C) 25.6 ×10-15 N
  • (D) 19.2 ×10-15 N
Show Answer

Answer: A – F = qvB = (1.60 × 10-19)(2×106)(0.04) = 12.8 ×10-15 N.

Q43. A straight wire of length 0.2 m carries a current of 4 A perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the wire?

  • (A) 0.32 N
  • (B) 0.16 N
  • (C) 0.08 N
  • (D) 0.24 N
Show Answer

Answer: B – F = ILB sinθ = (4)(0.2)(0.2) = 0.16 N.

Q44. A long straight wire carries a current of 12 A. What is the magnitude of the magnetic field at a distance of 0.05 m from the wire?

  • (A) 96 μT
  • (B) 24 μT
  • (C) 48 μT
  • (D) 72 μT
Show Answer

Answer: C – B = μ0I/(2πr) = (2×10-7)(12)/(0.05) = 48 μT.

Q45. A circular loop of radius 0.05 m carries a current of 5 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 126 μT
  • (B) 31.4 μT
  • (C) 62.8 μT
  • (D) 94.2 μT
Show Answer

Answer: C – B = μ0I/(2R) = (4π×10-7)(5)/(2×0.05) = 62.8 μT.

Q46. A solenoid has 400 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) 3.02 mT
  • (B) 0.754 mT
  • (C) 2.26 mT
  • (D) 1.51 mT
Show Answer

Answer: D – B = μ0nI = (4π×10-7)(400)(3) = 1.51 mT.

Q47. A proton moves at 15×104 m/s perpendicular to a uniform magnetic field of 0.05 T. What is the radius of its circular path?

  • (A) 6.26 cm
  • (B) 3.13 cm
  • (C) 1.57 cm
  • (D) 4.7 cm
Show Answer

Answer: B – qvB = mv²/r, so r = mv/(qB) = (1.67×10-27)(15×104)/((1.60×10-19)(0.05)) = 3.13 cm.

Q48. A rectangular coil of 20 turns, of sides 0.1 m by 0.1 m, carries a current of 3 A in a uniform magnetic field of 0.1 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) 60 mN·m
  • (B) 15 mN·m
  • (C) 30 mN·m
  • (D) 45 mN·m
Show Answer

Answer: C – τ = NIAB sinθ = (20)(3)(0.1×0.1)(0.1) sin 30° = 30 mN·m.

Q49. Two long parallel wires carry currents of 3 A and 6 A in the same direction and are 0.1 m apart. What is the magnitude of the magnetic force per unit length between the wires?

  • (A) 0.072 mN/m
  • (B) 0.018 mN/m
  • (C) 0.054 mN/m
  • (D) 0.036 mN/m
Show Answer

Answer: D – F/L = μ0I1I2/(2πd) = (2×10-7)(3)(6)/(0.1) = 0.036 mN/m. The wires attract each other.

Q50. A long straight cylindrical wire of radius 0.05 m carries a uniformly distributed current of 10 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) 16 μT
  • (B) 4 μT
  • (C) 8 μT
  • (D) 12 μT
Show Answer

Answer: C – By Ampere’s law inside the wire, B = μ0Ir/(2πR²) = (2×10-7)(10)(0.01)/(0.05)² = 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 0°. What is the magnetic flux through the loop?

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

Answer: C – Φ = BA cosθ = (0.15)(100×10-4) cos 0° = 1.5 mWb.

Q52. A coil of 40 turns and area 150 cm² is placed with its plane perpendicular to a magnetic field. The field strength increases uniformly at 3 T/s. What is the magnitude of the induced emf in the coil?

  • (A) 0.9×103 mV
  • (B) 3.6×103 mV
  • (C) 1.8×103 mV
  • (D) 2.7×103 mV
Show Answer

Answer: C – ε = NA(dB/dt) = (40)(150×10-4)(3) = 1.8×103 mV.

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 0.1 T. What is the magnitude of the motional emf between the ends of the rod?

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

Answer: B – ε = BLv = (0.1)(1)(10) = 1 V.

Q54. A coil with 1 turn, area 200 cm² and resistance 40 Ω lies perpendicular to a magnetic field that changes at 0.5 T/s. What is the magnitude of the induced current in the coil?

  • (A) 0.5 mA
  • (B) 0.125 mA
  • (C) 0.25 mA
  • (D) 0.375 mA
Show Answer

Answer: C – ε = N A(dB/dt) = (1)(200×10-4)(0.5) V, so I = ε/R = 0.25 mA. By Lenz’s law its direction opposes the change in flux.

Q55. An inductor of inductance 20 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) 0.8 V
  • (B) 0.2 V
  • (C) 0.6 V
  • (D) 0.4 V
Show Answer

Answer: D – ε = L|dI/dt| = (20×10-3)(20) = 0.4 V, opposing the change in current.

Q56. An inductor of inductance 50 mH carries a steady current of 2 A. How much energy is stored in its magnetic field?

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

Answer: C – U = (1/2)LI² = (1/2)(50×10-3)(2)² = 100 mJ.

Q57. An inductor of inductance 100 mH is connected in series with a resistor of 10 Ω. What is the time constant of this LR circuit?

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

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

Q58. A series LR circuit with a 9 V battery and a resistor of 10 Ω is switched on at t = 0 with zero initial current. The time constant is τ. What is the current at t = 1τ?

  • (A) 1.14 A
  • (B) 0.569 A
  • (C) 0.284 A
  • (D) 0.853 A
Show Answer

Answer: B – Solving L(dI/dt) + RI = V gives I(t) = (V/R)(1 – e-t/τ). With V/R = 0.9 A, at t = 1τ, I = 0.569 A.

Q59. An LC circuit consists of an inductor of 10 mH and a capacitor of 2 μF. What is the frequency of oscillation of the circuit?

  • (A) 2.25 kHz
  • (B) 0.563 kHz
  • (C) 1.13 kHz
  • (D) 1.69 kHz
Show Answer

Answer: C – f = 1/(2π√(LC)) = 1/(2π√[(10×10-3)(2×10-6)]) = 1.13 kHz.

Q60. An LC circuit has inductance 10 mH and capacitance 4 μF. What is the angular frequency of its charge oscillations?

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

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

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