AP Physics C: Electricity and Magnetism Practice Test 14 – 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 point charges of 2 μC and 6 μC are held 0.6 m apart in a vacuum. What is the magnitude of the electrostatic force exerted on either charge?

  • (A) 0.3 N
  • (B) 0.6 N
  • (C) 0.15 N
  • (D) 1.2 N
Show Answer

Answer: A – Use Coulomb’s law F = k q1 q2 / r2 = (9.0 × 109)(2 × 10-6)(6 × 10-6) / (0.6)2 = 0.3 N.

Q2. A point charge of 6 μC is fixed in a vacuum. What is the magnitude of the electric field at a point 0.2 m from the charge?

  • (A) 2.70 × 106 N/C
  • (B) 1.35 × 106 N/C
  • (C) 6.75 × 105 N/C
  • (D) 1.35 × 107 N/C
Show Answer

Answer: B – E = k q / r2 = (9.0 × 109)(6 × 10-6) / (0.2)2 = 1.35 × 106 N/C.

Q3. Two point charges, +8 μC and +6 μC, are fixed 0.4 m apart. What is the magnitude of the net electric field at the midpoint between them?

  • (A) 9.00 × 105 N/C
  • (B) 2.25 × 105 N/C
  • (C) 4.50 × 105 N/C
  • (D) 1.35 × 106 N/C
Show Answer

Answer: C – At the midpoint each field has magnitude k q / (0.2)2. Like charges give fields in opposite directions there, so Enet = k |q1 – q2| / (0.2)2 = 4.50 × 105 N/C.

Q4. A closed Gaussian surface encloses a net charge of 10 μC and no other charge. What is the net electric flux through the surface?

  • (A) 2.26 × 106 N·m2/C
  • (B) 1.13 × 106 N·m2/C
  • (C) 5.65 × 105 N·m2/C
  • (D) 1.13 × 107 N·m2/C
Show Answer

Answer: B – By Gauss’s law, Φ = Qenc / ε0 = (10 × 10-6) / (8.85 × 10-12) = 1.13 × 106 N·m2/C.

Q5. A solid insulating sphere of radius 0.1 m carries a total charge of 6 μC uniformly distributed through its volume. What is the magnitude of the electric field at a distance 0.3 m from the centre of the sphere?

  • (A) 6.00 × 105 N/C
  • (B) 1.20 × 106 N/C
  • (C) 3.00 × 105 N/C
  • (D) 2.40 × 106 N/C
Show Answer

Answer: A – Outside a spherically symmetric distribution, E = k Q / r2 = (9.0 × 109)(6 × 10-6) / (0.3)2 = 6.00 × 105 N/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 inside the sphere at a distance 0.05 m from its centre?

  • (A) 9.00 × 105 N/C
  • (B) 2.25 × 105 N/C
  • (C) 1.80 × 106 N/C
  • (D) 4.50 × 105 N/C
Show Answer

Answer: D – Inside a uniformly charged solid sphere, Gauss’s law gives E = k Q r / R3 = (9.0 × 109)(8 × 10-6)(0.05) / (0.2)3 = 4.50 × 105 N/C.

Q7. An infinitely long straight line of charge has a uniform linear charge density of 20 nC/m. What is the magnitude of the electric field at a perpendicular distance of 0.25 m from the line?

  • (A) 2880 N/C
  • (B) 1440 N/C
  • (C) 720 N/C
  • (D) 5760 N/C
Show Answer

Answer: B – For an infinite line, E = 2kλ / r = 2(9.0 × 109)(20 × 10-9) / (0.25) = 1440 N/C.

Q8. An infinite nonconducting sheet carries a uniform surface charge density of 6 μC/m2. What is the magnitude of the electric field near the sheet, on either side of it?

  • (A) 339 kN/C
  • (B) 678 kN/C
  • (C) 169 kN/C
  • (D) 1360 kN/C
Show Answer

Answer: A – For a single infinite sheet, E = σ / (2ε0) = (6 × 10-6) / (2 × 8.85 × 10-12) = 339 kN/C.

Q9. A thin ring of radius 0.4 m carries a total charge of 6 μC uniformly distributed around it. What is the magnitude of the electric field on the axis of the ring at a distance of 0.3 m from its centre?

  • (A) 1.30 × 105 N/C
  • (B) 2.59 × 105 N/C
  • (C) 6.48 × 104 N/C
  • (D) 1.30 × 106 N/C
Show Answer

Answer: A – On the axis, E = k Q x / (R2 + x2)3/2 = (9.0 × 109)(6 × 10-6)(0.3) / (0.125) = 1.30 × 105 N/C.

Q10. A uniform electric field of 200 N/C passes through a flat surface of area 0.04 m2. The angle between the field and the normal to the surface is 0°. What is the electric flux through the surface?

  • (A) 16 N·m2/C
  • (B) 4 N·m2/C
  • (C) 32 N·m2/C
  • (D) 8 N·m2/C
Show Answer

Answer: D – Φ = E A cos θ = (200)(0.04) cos 0° = 8 N·m2/C.

Q11. The electric field at the surface of a uniformly charged spherical shell has magnitude 4000 N/C, directed radially outward, and the shell has radius 0.25 m. What is the total charge on the shell?

  • (A) 0.0556 μC
  • (B) 0.0139 μC
  • (C) 0.111 μC
  • (D) 0.0278 μC
Show Answer

Answer: D – Outside the shell E = k Q / r2, so Q = E r2 / k = (4000)(0.25)2 / (9.0 × 109) = 0.0278 μC.

Q12. A thin rod of length 0.5 m carries a uniform linear charge density λ. A point P lies on the axis of the rod at a distance 0.2 m from its nearer end. Which integral gives the magnitude of the electric field at P?

  • (A) ∪00.5 kλ dx / (0.2 + x)
  • (B) ∪00.5 kλ dx / (0.2 + x)2
  • (C) ∪00.5 kλ x dx / (0.2 + x)2
  • (D) kλ0.5 / (0.2)2
Show Answer

Answer: B – An element dq = λ dx at distance (0.2 + x) from P contributes dE = k dq / r2 = kλ dx / (0.2 + x)2. Integrating from x = 0 to x = 0.5 gives the field.

Unit 9: Electric Potential

Q13. A point charge of 3 μC is fixed in a vacuum. Taking the potential at infinity to be zero, what is the electric potential at a distance 0.3 m from the charge?

  • (A) 180 kV
  • (B) 45 kV
  • (C) 90 kV
  • (D) 360 kV
Show Answer

Answer: C – V = k q / r = (9.0 × 109)(3 × 10-6) / (0.3) = 90 kV.

Q14. A charge of 4 μC is moved slowly from a point at potential 0 V to a point at potential 2000 V by an external agent. How much work is done by the external agent?

  • (A) 16 mJ
  • (B) 8 mJ
  • (C) 4 mJ
  • (D) 32 mJ
Show Answer

Answer: B – W = q ΔV = (4 × 10-6)(2000) = 8 mJ.

Q15. Two point charges of 4 μC and 4 μC are 0.2 m apart. What is the electric potential energy of this pair, taking zero energy at infinite separation?

  • (A) 1440 mJ
  • (B) 720 mJ
  • (C) 360 mJ
  • (D) 2880 mJ
Show Answer

Answer: B – U = k q1 q2 / r = (9.0 × 109)(4 × 10-6)(4 × 10-6) / (0.2) = 720 mJ.

Q16. A thin ring of radius 0.3 m carries a total charge of 8 μC uniformly distributed around it. What is the electric potential on the axis of the ring at a distance of 0.3 m from its centre, with zero potential at infinity?

  • (A) 339 kV
  • (B) 170 kV
  • (C) 84.9 kV
  • (D) 679 kV
Show Answer

Answer: B – Every charge element is 0.424 m from the point, so V = k Q / √(R2 + x2) = (9.0 × 109)(8 × 10-6) / (0.424) = 170 kV.

Q17. A thin rod of length 0.6 m carries a uniform linear charge density λ. A point P lies on the axis of the rod at a distance 0.2 m from its nearer end. Which integral gives the electric potential at P, with zero potential at infinity?

  • (A) ∪00.6 kλ dx / (0.2 + x)2
  • (B) ∪00.6 kλ dx / (0.2 + x)
  • (C) ∪00.6 kλ x dx / (0.2 + x)
  • (D) kλ0.6 / (0.2)
Show Answer

Answer: B – An element dq = λ dx contributes dV = k dq / r = kλ dx / (0.2 + x). Integrating from x = 0 to x = 0.6 gives V at P.

Q18. In a region, the electric potential varies with position as V(x) = 4x2 + 6x + 5, where V is in volts and x is in metres. What is the x-component of the electric field at x = 3 m?

  • (A) -30 V/m
  • (B) -60 V/m
  • (C) -15 V/m
  • (D) 30 V/m
Show Answer

Answer: A – Ex = -dV/dx = -(8x + 6). At x = 3 m, Ex = -(8 × 3 + 6) = -30 V/m.

Q19. A conducting spherical shell of radius 0.2 m carries a total charge of 6 μC. What is the electric potential at a point inside the shell, 0.05 m from its centre, with zero potential at infinity?

  • (A) 540 kV
  • (B) 135 kV
  • (C) 270 kV
  • (D) 1080 kV
Show Answer

Answer: C – Inside a conducting shell the potential is constant and equal to its surface value, V = k Q / R = (9.0 × 109)(6 × 10-6) / (0.2) = 270 kV.

Q20. A particle of charge 5 μC is released from rest at a point where the potential is 1000 V higher than at a second point. If no non-electric forces act, what kinetic energy does it gain in moving to the second point?

  • (A) 10 mJ
  • (B) 2.5 mJ
  • (C) 5 mJ
  • (D) 20 mJ
Show Answer

Answer: C – Loss of potential energy becomes kinetic energy: ΔK = q ΔV = (5 × 10-6)(1000) = 5 mJ.

Q21. An electric dipole consists of charges +4 μC and -4 μC separated by 0.04 m. Estimate the electric potential on the dipole axis at a distance 0.2 m from its centre, with zero potential at infinity, using the far-field dipole approximation.

  • (A) 72 kV
  • (B) 36 kV
  • (C) 18 kV
  • (D) 144 kV
Show Answer

Answer: B – p = q d = (4 × 10-6)(0.04) and V ≈ k p / r2 = (9.0 × 109)(1.60 × 10-7) / (0.2)2 = 36 kV.

Unit 10: Conductors and Capacitors

Q22. A parallel-plate capacitor has plates of area 0.01 m2 separated by 2 mm in a vacuum. What is its capacitance?

  • (A) 44.2 pF
  • (B) 88.5 pF
  • (C) 22.1 pF
  • (D) 177 pF
Show Answer

Answer: A – C = ε0 A / d = (8.85 × 10-12)(0.01) / (2 × 10-3) = 44.2 pF.

Q23. A parallel-plate capacitor has capacitance 100 pF in air. A dielectric slab of dielectric constant 3 is inserted to fill the space between the plates completely, with the plates isolated so charge stays fixed. What is the new capacitance?

  • (A) 600 pF
  • (B) 150 pF
  • (C) 1200 pF
  • (D) 300 pF
Show Answer

Answer: D – With a dielectric, C = κ C0 = (3)(100 pF) = 300 pF.

Q24. Capacitors of 9 μF and 6 μF are connected in series. What is their equivalent capacitance?

  • (A) 7.2 μF
  • (B) 1.8 μF
  • (C) 10.8 μF
  • (D) 3.6 μF
Show Answer

Answer: D – 1/Ceq = 1/9 + 1/6 = 1/3.6 μF-1, so Ceq = 3.6 μF.

Q25. A capacitor of 4 μF is charged to a potential difference of 300 V. How much energy is stored in it?

  • (A) 360 mJ
  • (B) 90 mJ
  • (C) 720 mJ
  • (D) 180 mJ
Show Answer

Answer: D – U = (1/2) C V2 = (1/2)(4 × 10-6)(300)2 = 180 mJ.

Q26. A capacitor of 4 μF is connected across a 100 V battery until fully charged. What charge is stored on either plate?

  • (A) 800 μC
  • (B) 200 μC
  • (C) 1600 μC
  • (D) 400 μC
Show Answer

Answer: D – Q = C V = (4 μF)(100 V) = 400 μC.

Q27. A spherical capacitor consists of concentric conducting spheres of radii 0.05 m and 0.2 m, with vacuum between them. What is its capacitance?

  • (A) 14.8 pF
  • (B) 3.71 pF
  • (C) 7.41 pF
  • (D) 29.7 pF
Show Answer

Answer: C – C = 4πε0 a b / (b – a) = 4π(8.85 × 10-12)(0.05)(0.2) / (0.15) = 7.41 pF.

Q28. A solid conducting sphere is given a net positive charge and allowed to reach electrostatic equilibrium. Which statement is correct?

  • (A) Excess charge is spread uniformly through the volume, and the field inside increases with radius.
  • (B) All excess charge resides at the centre, and the field inside is uniform and nonzero.
  • (C) Excess charge resides on the outer surface, and the field inside is the same as the field just outside.
  • (D) All excess charge resides on the outer surface, and the electric field inside the sphere is zero.
Show Answer

Answer: D – In electrostatic equilibrium free charges move until the field inside a conductor is zero, so excess charge sits on the outer surface.

Q29. The electric field between the plates of a capacitor is 25000 N/C. What is the energy density stored in the field?

  • (A) 5.53 mJ/m3
  • (B) 1.38 mJ/m3
  • (C) 11.1 mJ/m3
  • (D) 2.77 mJ/m3
Show Answer

Answer: D – u = (1/2) ε0 E2 = (1/2)(8.85 × 10-12)(25000)2 = 2.77 mJ/m3.

Unit 11: Electric Circuits

Q30. A wire of length 2 m and cross-sectional area 1 mm2 is made of a material of resistivity 2.8 × 10-8 Ω·m. What is the resistance of the wire?

  • (A) 0.056 Ω
  • (B) 0.112 Ω
  • (C) 0.028 Ω
  • (D) 0.224 Ω
Show Answer

Answer: A – R = ρ L / A = (2.8 × 10-8)(2) / (1 × 10-6) = 0.056 Ω.


Q31. Resistors of 6 Ω and 3 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 4 Ω
  • (B) 1 Ω
  • (C) 6 Ω
  • (D) 2 Ω
Show Answer

Answer: D – 1/Req = 1/6 + 1/3, so Req = 2 Ω.

Q32. A 12 V battery with negligible internal resistance is connected across a 4 Ω resistor. What current flows through the resistor?

  • (A) 6 A
  • (B) 1.5 A
  • (C) 3 A
  • (D) 12 A
Show Answer

Answer: C – I = V / R = 12 / 4 = 3 A.

Q33. In a single loop, two batteries of emfs 18 V and 8 V oppose each other, and the total resistance in the loop is 9 Ω. What is the current in the loop, in the direction of the larger emf?

  • (A) 2.22 A
  • (B) 1.11 A
  • (C) 0.556 A
  • (D) 4.44 A
Show Answer

Answer: B – Kirchhoff’s loop rule gives I = (E1 – E2) / R = (18 – 8) / 9 = 1.11 A.

Q34. A current of 2 A flows through a 10 Ω resistor. At what rate is energy dissipated as heat in the resistor?

  • (A) 80 W
  • (B) 20 W
  • (C) 40 W
  • (D) 160 W
Show Answer

Answer: C – P = I2 R = (2)2(10) = 40 W.

Q35. A capacitor of 100 μF is connected in series with a 5 kΩ resistor. What is the time constant of this RC circuit?

  • (A) 1 s
  • (B) 0.25 s
  • (C) 0.5 s
  • (D) 2 s
Show Answer

Answer: C – τ = R C = (5 × 103)(100 × 10-6) = 0.5 s.

Q36. A capacitor in an RC circuit charges toward a maximum charge of 60 μC. After a time equal to 2 time constant(s), what charge is on the capacitor?

  • (A) 104 μC
  • (B) 25.9 μC
  • (C) 51.9 μC
  • (D) 208 μC
Show Answer

Answer: C – q(t) = Qmax(1 – e-t/τ) = 60(1 – e-2) = 51.9 μC.

Q37. A discharging RC circuit has an initial current of 10 mA. After a time equal to 1 time constant(s), what is the current?

  • (A) 7.36 mA
  • (B) 1.84 mA
  • (C) 3.68 mA
  • (D) 14.7 mA
Show Answer

Answer: C – I(t) = I0 e-t/τ = 10 e-1 = 3.68 mA.

Q38. An RC circuit with time constant 0.5 s is charging from zero charge. How long does it take for the charge to reach 90% of its maximum value?

  • (A) 1.15 s
  • (B) 2.3 s
  • (C) 0.576 s
  • (D) 4.61 s
Show Answer

Answer: A – q/Qmax = 1 – e-t/τ = 0.9, so t = -τ ln(1 – 0.9) = 0.5 × 2.303 = 1.15 s.

Q39. A battery of emf 12 V and internal resistance 1 Ω delivers a current of 2 A to an external circuit. What is the terminal potential difference of the battery?

  • (A) 10 V
  • (B) 20 V
  • (C) 5 V
  • (D) 40 V
Show Answer

Answer: A – V = ε – I r = 12 – (2)(1) = 10 V.

Q40. Resistors of 6 Ω and 8 Ω are connected in parallel across a 24 V battery. What is the current through the 6 Ω resistor?

  • (A) 8 A
  • (B) 4 A
  • (C) 2 A
  • (D) 16 A
Show Answer

Answer: B – Each parallel branch sees the full battery voltage, so I1 = V / R1 = 24 / 6 = 4 A.

Q41. In a series circuit, a 24 V battery is connected to a 6 Ω resistor in series with a 6 Ω resistor, and a capacitor is connected in parallel with the 6 Ω resistor. In steady state, what is the potential difference across the capacitor?

  • (A) 24 V
  • (B) 6 V
  • (C) 48 V
  • (D) 12 V
Show Answer

Answer: D – In steady state no current flows through the capacitor branch, so the capacitor voltage equals the voltage across the 6 Ω resistor: VC = V R2 / (R1 + R2) = 24 × 6 / 12 = 12 V.

Unit 12: Magnetic Fields and Electromagnetism

Q42. A particle of charge 5 μC moves at 3 × 104 m/s perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the particle?

  • (A) 60 mN
  • (B) 15 mN
  • (C) 120 mN
  • (D) 30 mN
Show Answer

Answer: D – F = q v B sin 90° = (5 × 10-6)(3 × 104)(0.2) = 30 mN.

Q43. An alpha particle (mass 6.4 × 10-27 kg, charge 3.2 × 10-19 C) moves at 6 × 106 m/s perpendicular to a uniform magnetic field of 1 T. What is the radius of its circular path?

  • (A) 12 cm
  • (B) 24 cm
  • (C) 6 cm
  • (D) 48 cm
Show Answer

Answer: A – r = m v / (q B) = (6.4 × 10-27)(6 × 106) / ((3.2 × 10-19)(1)) = 12 cm.

Q44. A straight wire of length 1.5 m carries a current of 4 A perpendicular to a uniform magnetic field of 20 mT. What is the magnitude of the magnetic force on the wire?

  • (A) 120 mN
  • (B) 240 mN
  • (C) 60 mN
  • (D) 480 mN
Show Answer

Answer: A – F = I L B = (4)(1.5)(20 × 10-3) = 120 mN.

Q45. A long straight wire carries a current of 5 A. What is the magnitude of the magnetic field at a distance of 10 cm from the wire?

  • (A) 20 μT
  • (B) 5 μT
  • (C) 10 μT
  • (D) 40 μT
Show Answer

Answer: C – B = μ0 I / (2π r) = (4π × 10-7)(5) / (2π × 0.1) = 10 μT.

Q46. A circular loop of radius 10 cm carries a current of 2 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 25.1 μT
  • (B) 12.6 μT
  • (C) 6.28 μT
  • (D) 50.3 μT
Show Answer

Answer: B – B = μ0 I / (2R) = (4π × 10-7)(2) / (2 × 0.1) = 12.6 μT.

Q47. A long solenoid has 1500 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) 5.65 mT
  • (B) 11.3 mT
  • (C) 2.83 mT
  • (D) 22.6 mT
Show Answer

Answer: A – B = μ0 n I = (4π × 10-7)(1500)(3) = 5.65 mT.

Q48. Two long parallel wires 20 cm apart carry currents of 2 A and 4 A in opposite directions. What is the magnitude of the net magnetic field at the midpoint between the wires?

  • (A) 24 μT
  • (B) 6 μT
  • (C) 48 μT
  • (D) 12 μT
Show Answer

Answer: D – At the midpoint each wire contributes B = μ0 I / (2π r) with r = 0.1 m, and for opposite currents the fields point the same way, so Bnet = μ0(I1 + I2) / (2π r) = 12 μT.

Q49. A rectangular coil of 20 turns and area 0.04 m2 carries a current of 2 A in a uniform magnetic field of 0.2 T. When the plane of the coil is parallel to the field, what is the magnitude of the torque on the coil?

  • (A) 640 mN·m
  • (B) 320 mN·m
  • (C) 160 mN·m
  • (D) 1280 mN·m
Show Answer

Answer: B – Maximum torque τ = N I A B = (20)(2)(0.04)(0.2) = 320 mN·m.

Q50. In a velocity selector, a charged particle passes undeflected through crossed electric and magnetic fields of 4000 N/C and 20 mT. What is the speed of the particle?

  • (A) 4.00 × 105 m/s
  • (B) 2.00 × 105 m/s
  • (C) 1.00 × 105 m/s
  • (D) 8.00 × 105 m/s
Show Answer

Answer: B – For no deflection qE = qvB, so v = E / B = 4000 / (20 × 10-3) = 2.00 × 105 m/s.

Unit 13: Electromagnetic Induction

Q51. A flat loop of area 0.04 m2 is placed in a uniform magnetic field of 100 mT. The angle between the field and the normal to the loop is 0°. What is the magnetic flux through the loop?

  • (A) 8 mWb
  • (B) 2 mWb
  • (C) 4 mWb
  • (D) 16 mWb
Show Answer

Answer: C – Φ = B A cos θ = (100 × 10-3)(0.04) cos 0° = 4 mWb.

Q52. The magnetic field through a single loop of area 0.1 m2 increases uniformly by 0.5 T in 0.2 s, with the field perpendicular to the loop. What is the magnitude of the induced emf in the loop?

  • (A) 500 mV
  • (B) 125 mV
  • (C) 250 mV
  • (D) 1000 mV
Show Answer

Answer: C – |ε| = A ΔB / Δt = (0.1)(0.5) / (0.2) = 250 mV.

Q53. A conducting rod of length 1 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.2 T, with the rod perpendicular to both its velocity and the field. What is the motional emf between the ends of the rod?

  • (A) 0.8 V
  • (B) 0.2 V
  • (C) 1.6 V
  • (D) 0.4 V
Show Answer

Answer: D – ε = B L v = (0.2)(1)(2) = 0.4 V.

Q54. A conducting loop lies in the plane of the page in a magnetic field directed out of the page. The field strength decreases with time. Viewed from above the page, in which direction does the induced current flow?

  • (A) Counterclockwise, because the induced field must point out of the page to oppose the decreasing out-of-page flux.
  • (B) Clockwise, because the induced field must point out of the page to oppose the decreasing out-of-page flux.
  • (C) Counterclockwise, because the induced field must point into the page to speed the decrease.
  • (D) Clockwise, because a decreasing field always induces a clockwise current.
Show Answer

Answer: A – Decreasing out-of-page flux requires an induced field in the original direction, out of the page, which corresponds to a counterclockwise current.

Q55. A solenoid has 400 turns, a cross-sectional area of 40 cm2, and a length of 0.2 m. What is its self-inductance?

  • (A) 8.04 mH
  • (B) 2.01 mH
  • (C) 4.02 mH
  • (D) 16.1 mH
Show Answer

Answer: C – L = μ0 N2 A / l = (4π × 10-7)(400)2(40 × 10-4) / (0.2) = 4.02 mH.

Q56. An inductor of 100 mH is connected in series with a 5 Ω resistor. What is the time constant of this LR circuit?

  • (A) 40 ms
  • (B) 10 ms
  • (C) 20 ms
  • (D) 80 ms
Show Answer

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

Q57. A series LR circuit with battery voltage 24 V and resistance 12 Ω is switched on at t = 0. After a time equal to 2 time constant(s), what is the current in the circuit?

  • (A) 3.46 A
  • (B) 0.865 A
  • (C) 6.92 A
  • (D) 1.73 A
Show Answer

Answer: D – I(t) = (V/R)(1 – e-t/τ) = (24/12)(1 – e-2) = 1.73 A.

Q58. An LC circuit consists of an inductor of 40 mH and a capacitor of 40 μF. What is its natural frequency of oscillation?

  • (A) 252 Hz
  • (B) 126 Hz
  • (C) 62.9 Hz
  • (D) 503 Hz
Show Answer

Answer: B – f = 1 / (2π√(LC)) = 1 / (2π√((40 × 10-3)(40 × 10-6))) = 126 Hz.

Q59. In an ideal LC circuit, a capacitor of 40 μF is initially charged to 20 V and then connected to an inductor of 10 mH. What is the maximum current in the circuit?

  • (A) 1.26 A
  • (B) 2.53 A
  • (C) 0.632 A
  • (D) 5.06 A
Show Answer

Answer: A – Energy conservation gives (1/2) C V2 = (1/2) L Imax2, so Imax = V √(C/L) = 20 √((40 × 10-6) / (10 × 10-3)) = 1.26 A.

Q60. A coil of 20 turns and area 0.04 m2 rotates in a uniform magnetic field of 0.5 T with angular speed 20 rad/s. What is the maximum induced emf in the coil?

  • (A) 8 V
  • (B) 16 V
  • (C) 4 V
  • (D) 32 V
Show Answer

Answer: A – εmax = N B A ω = (20)(0.5)(0.04)(20) = 8 V.

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