AP Physics C: Electricity and Magnetism Practice Test 13 – 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 0.4 m apart in a vacuum. What is the magnitude of the electrostatic force exerted on either charge?

  • (A) 0.675 N
  • (B) 1.35 N
  • (C) 0.337 N
  • (D) 2.7 N
Show Answer

Answer: A – Use Coulomb’s law F = k q1 q2 / r2 = (9.0 × 109)(2 × 10-6)(6 × 10-6) / (0.4)2 = 0.675 N.

Q2. A point charge of 8 μC is fixed in a vacuum. What is the magnitude of the electric field at a point 0.2 m from the charge?

  • (A) 3.60 × 106 N/C
  • (B) 1.80 × 106 N/C
  • (C) 9.00 × 105 N/C
  • (D) 1.80 × 107 N/C
Show Answer

Answer: B – E = k q / r2 = (9.0 × 109)(8 × 10-6) / (0.2)2 = 1.80 × 106 N/C.

Q3. Two point charges, +2 μC and +8 μC, are fixed 0.8 m apart. What is the magnitude of the net electric field at the midpoint between them?

  • (A) 6.75 × 105 N/C
  • (B) 1.69 × 105 N/C
  • (C) 3.37 × 105 N/C
  • (D) 1.01 × 106 N/C
Show Answer

Answer: C – At the midpoint each field has magnitude k q / (0.4)2. Like charges give fields in opposite directions there, so Enet = k |q1 – q2| / (0.4)2 = 3.37 × 105 N/C.

Q4. A closed Gaussian surface encloses a net charge of 8 μC and no other charge. What is the net electric flux through the surface?

  • (A) 1.81 × 106 N·m2/C
  • (B) 4.52 × 105 N·m2/C
  • (C) 9.04 × 106 N·m2/C
  • (D) 9.04 × 105 N·m2/C
Show Answer

Answer: D – By Gauss’s law, Φ = Qenc / ε0 = (8 × 10-6) / (8.85 × 10-12) = 9.04 × 105 N·m2/C.

Q5. A solid insulating sphere of radius 0.2 m carries a total charge of 4 μC uniformly distributed through its volume. What is the magnitude of the electric field at a distance 0.5 m from the centre of the sphere?

  • (A) 1.44 × 105 N/C
  • (B) 2.88 × 105 N/C
  • (C) 7.20 × 104 N/C
  • (D) 5.76 × 105 N/C
Show Answer

Answer: A – Outside a spherically symmetric distribution, E = k Q / r2 = (9.0 × 109)(4 × 10-6) / (0.5)2 = 1.44 × 105 N/C.

Q6. A solid insulating sphere of radius 0.3 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 0.075 m from its centre?

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

Answer: C – Inside a uniformly charged solid sphere, Gauss’s law gives E = k Q r / R3 = (9.0 × 109)(8 × 10-6)(0.075) / (0.3)3 = 2.00 × 105 N/C.

Q7. An infinitely long straight line of charge has a uniform linear charge density of 5 nC/m. What is the magnitude of the electric field at a perpendicular distance of 0.25 m from the line?

  • (A) 720 N/C
  • (B) 180 N/C
  • (C) 360 N/C
  • (D) 1440 N/C
Show Answer

Answer: C – For an infinite line, E = 2kλ / r = 2(9.0 × 109)(5 × 10-9) / (0.25) = 360 N/C.

Q8. An infinite nonconducting sheet carries a uniform surface charge density of 3 μC/m2. What is the magnitude of the electric field near the sheet, on either side of it?

  • (A) 339 kN/C
  • (B) 84.7 kN/C
  • (C) 678 kN/C
  • (D) 169 kN/C
Show Answer

Answer: D – For a single infinite sheet, E = σ / (2ε0) = (3 × 10-6) / (2 × 8.85 × 10-12) = 169 kN/C.

Q9. A thin ring of radius 0.4 m carries a total charge of 8 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) 1.73 × 105 N/C
  • (B) 3.46 × 105 N/C
  • (C) 8.64 × 104 N/C
  • (D) 1.73 × 106 N/C
Show Answer

Answer: A – On the axis, E = k Q x / (R2 + x2)3/2 = (9.0 × 109)(8 × 10-6)(0.3) / (0.125) = 1.73 × 105 N/C.

Q10. A uniform electric field of 600 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) 48 N·m2/C
  • (B) 12 N·m2/C
  • (C) 24 N·m2/C
  • (D) 96 N·m2/C
Show Answer

Answer: C – Φ = E A cos θ = (600)(0.04) cos 0° = 24 N·m2/C.

Q11. The electric field at the surface of a uniformly charged spherical shell has magnitude 2000 N/C, directed radially outward, and the shell has radius 0.1 m. What is the total charge on the shell?

  • (A) 4.44 × 10-3 μC
  • (B) 1.11 × 10-3 μC
  • (C) 2.22 × 10-3 μC
  • (D) 8.89 × 10-3 μC
Show Answer

Answer: C – Outside the shell E = k Q / r2, so Q = E r2 / k = (2000)(0.1)2 / (9.0 × 109) = 2.22 × 10-3 μC.

Q12. A thin rod of length 0.45 m carries a uniform linear charge density λ. A point P lies on the axis of the rod at a distance 0.15 m from its nearer end. Which integral gives the magnitude of the electric field at P?

  • (A) ∪00.45 kλ dx / (0.15 + x)2
  • (B) ∪00.45 kλ dx / (0.15 + x)
  • (C) ∪00.45 kλ x dx / (0.15 + x)2
  • (D) kλ0.45 / (0.15)2
Show Answer

Answer: A – An element dq = λ dx at distance (0.15 + x) from P contributes dE = k dq / r2 = kλ dx / (0.15 + x)2. Integrating from x = 0 to x = 0.45 gives the field.

Unit 9: Electric Potential

Q13. A point charge of 8 μC is fixed in a vacuum. Taking the potential at infinity to be zero, what is the electric potential at a distance 0.4 m from the charge?

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

Answer: C – V = k q / r = (9.0 × 109)(8 × 10-6) / (0.4) = 180 kV.

Q14. A charge of 2 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) 8 mJ
  • (B) 4 mJ
  • (C) 2 mJ
  • (D) 16 mJ
Show Answer

Answer: B – W = q ΔV = (2 × 10-6)(2000) = 4 mJ.

Q15. Two point charges of 3 μC and 3 μC are 0.3 m apart. What is the electric potential energy of this pair, taking zero energy at infinite separation?

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

Answer: D – U = k q1 q2 / r = (9.0 × 109)(3 × 10-6)(3 × 10-6) / (0.3) = 270 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 0.4 m from its centre, with zero potential at infinity?

  • (A) 288 kV
  • (B) 72 kV
  • (C) 576 kV
  • (D) 144 kV
Show Answer

Answer: D – Every charge element is 0.5 m from the point, so V = k Q / √(R2 + x2) = (9.0 × 109)(8 × 10-6) / (0.5) = 144 kV.

Q17. A thin rod of length 0.55 m carries a uniform linear charge density λ. A point P lies on the axis of the rod at a distance 0.25 m from its nearer end. Which integral gives the electric potential at P, with zero potential at infinity?

  • (A) ∪00.55 kλ dx / (0.25 + x)2
  • (B) ∪00.55 kλ dx / (0.25 + x)
  • (C) ∪00.55 kλ x dx / (0.25 + x)
  • (D) kλ0.55 / (0.25)
Show Answer

Answer: B – An element dq = λ dx contributes dV = k dq / r = kλ dx / (0.25 + x). Integrating from x = 0 to x = 0.55 gives V at P.

Q18. In a region, the electric potential varies with position as V(x) = 4x2 + 10x + 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) -68 V/m
  • (B) -34 V/m
  • (C) -17 V/m
  • (D) 34 V/m
Show Answer

Answer: B – Ex = -dV/dx = -(8x + 10). At x = 3 m, Ex = -(8 × 3 + 10) = -34 V/m.

Q19. A conducting spherical shell of radius 0.25 m carries a total charge of 4 μC. What is the electric potential at a point inside the shell, 0.05 m from its centre, with zero potential at infinity?

  • (A) 288 kV
  • (B) 72 kV
  • (C) 144 kV
  • (D) 576 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)(4 × 10-6) / (0.25) = 144 kV.

Q20. A particle of charge 2 μC is released from rest at a point where the potential is 2000 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) 4 mJ
  • (B) 8 mJ
  • (C) 2 mJ
  • (D) 16 mJ
Show Answer

Answer: A – Loss of potential energy becomes kinetic energy: ΔK = q ΔV = (2 × 10-6)(2000) = 4 mJ.

Q21. An electric dipole consists of charges +2 μC and -2 μC separated by 0.02 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) 18 kV
  • (B) 4.5 kV
  • (C) 9 kV
  • (D) 36 kV
Show Answer

Answer: C – p = q d = (2 × 10-6)(0.02) and V ≈ k p / r2 = (9.0 × 109)(4.00 × 10-8) / (0.2)2 = 9 kV.

Unit 10: Conductors and Capacitors

Q22. A parallel-plate capacitor has plates of area 0.02 m2 separated by 4 mm in a vacuum. What is its capacitance?

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

Answer: B – C = ε0 A / d = (8.85 × 10-12)(0.02) / (4 × 10-3) = 44.2 pF.

Q23. A parallel-plate capacitor has capacitance 40 pF in air. A dielectric slab of dielectric constant 2 is inserted to fill the space between the plates completely, with the plates isolated so charge stays fixed. What is the new capacitance?

  • (A) 160 pF
  • (B) 80 pF
  • (C) 40 pF
  • (D) 320 pF
Show Answer

Answer: B – With a dielectric, C = κ C0 = (2)(40 pF) = 80 pF.

Q24. Capacitors of 12 μF and 6 μF are connected in series. What is their equivalent capacitance?

  • (A) 4 μF
  • (B) 8 μF
  • (C) 2 μF
  • (D) 12 μF
Show Answer

Answer: A – 1/Ceq = 1/12 + 1/6 = 1/4 μF-1, so Ceq = 4 μF.

Q25. A capacitor of 2 μF is charged to a potential difference of 300 V. How much energy is stored in it?

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

Answer: A – U = (1/2) C V2 = (1/2)(2 × 10-6)(300)2 = 90 mJ.

Q26. A capacitor of 10 μF is connected across a 50 V battery until fully charged. What charge is stored on either plate?

  • (A) 1000 μC
  • (B) 250 μC
  • (C) 500 μC
  • (D) 2000 μC
Show Answer

Answer: C – Q = C V = (10 μF)(50 V) = 500 μC.

Q27. A spherical capacitor consists of concentric conducting spheres of radii 0.05 m and 0.15 m, with vacuum between them. What is its capacitance?

  • (A) 16.7 pF
  • (B) 8.34 pF
  • (C) 4.17 pF
  • (D) 33.4 pF
Show Answer

Answer: B – C = 4πε0 a b / (b – a) = 4π(8.85 × 10-12)(0.05)(0.15) / (0.1) = 8.34 pF.

Q28. A hollow conducting shell is given a net positive charge and allowed to reach electrostatic equilibrium, with no charge in its cavity. Which statement is correct?

  • (A) Excess charge splits equally between the inner and outer surfaces, and the field in the cavity is nonzero.
  • (B) All excess charge resides on the outer surface of the shell, and the field in the cavity is zero.
  • (C) All excess charge resides on the inner surface, and the field outside the shell is zero.
  • (D) Excess charge resides on the outer surface, and the field in the cavity equals the field just outside.
Show Answer

Answer: B – With no charge in the cavity, Gauss’s law gives zero field in the cavity and all excess charge on the outer surface.

Q29. The electric field between the plates of a capacitor is 50000 N/C. What is the energy density stored in the field?

  • (A) 22.1 mJ/m3
  • (B) 11.1 mJ/m3
  • (C) 5.53 mJ/m3
  • (D) 44.2 mJ/m3
Show Answer

Answer: B – u = (1/2) ε0 E2 = (1/2)(8.85 × 10-12)(50000)2 = 11.1 mJ/m3.

Unit 11: Electric Circuits

Q30. A wire of length 4 m and cross-sectional area 1 mm2 is made of a material of resistivity 5.0 × 10-8 Ω·m. What is the resistance of the wire?

  • (A) 0.4 Ω
  • (B) 0.1 Ω
  • (C) 0.8 Ω
  • (D) 0.2 Ω
Show Answer

Answer: D – R = ρ L / A = (5.0 × 10-8)(4) / (1 × 10-6) = 0.2 Ω.


Q31. Resistors of 10 Ω and 15 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.

Q32. A 12 V battery with negligible internal resistance is connected across a 6 Ω resistor. What current flows through the resistor?

  • (A) 4 A
  • (B) 2 A
  • (C) 1 A
  • (D) 8 A
Show Answer

Answer: B – I = V / R = 12 / 6 = 2 A.

Q33. In a single loop, two batteries of emfs 12 V and 6 V oppose each other, and the total resistance in the loop is 6 Ω. What is the current in the loop, in the direction of the larger emf?

  • (A) 1 A
  • (B) 2 A
  • (C) 0.5 A
  • (D) 4 A
Show Answer

Answer: A – Kirchhoff’s loop rule gives I = (E1 – E2) / R = (12 – 6) / 6 = 1 A.

Q34. A current of 4 A flows through a 5 Ω resistor. At what rate is energy dissipated as heat in the resistor?

  • (A) 80 W
  • (B) 160 W
  • (C) 40 W
  • (D) 320 W
Show Answer

Answer: A – P = I2 R = (4)2(5) = 80 W.

Q35. A capacitor of 50 μF is connected in series with a 10 kΩ resistor. What is the time constant of this RC circuit?

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

Answer: D – τ = R C = (10 × 103)(50 × 10-6) = 0.5 s.

Q36. A capacitor in an RC circuit charges toward a maximum charge of 20 μC. After a time equal to 1 time constant(s), what charge is on the capacitor?

  • (A) 25.3 μC
  • (B) 6.32 μC
  • (C) 12.6 μC
  • (D) 50.6 μC
Show Answer

Answer: C – q(t) = Qmax(1 – e-t/τ) = 20(1 – e-1) = 12.6 μC.

Q37. A discharging RC circuit has an initial current of 20 mA. After a time equal to 2 time constant(s), what is the current?

  • (A) 2.71 mA
  • (B) 5.41 mA
  • (C) 1.35 mA
  • (D) 10.8 mA
Show Answer

Answer: A – I(t) = I0 e-t/τ = 20 e-2 = 2.71 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 75% of its maximum value?

  • (A) 1.39 s
  • (B) 0.346 s
  • (C) 2.77 s
  • (D) 0.693 s
Show Answer

Answer: D – q/Qmax = 1 – e-t/τ = 0.75, so t = -τ ln(1 – 0.75) = 0.5 × 1.386 = 0.693 s.

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

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

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

Q40. Resistors of 6 Ω and 12 Ω are connected in parallel across a 12 V battery. What is the current through the 6 Ω resistor?

  • (A) 2 A
  • (B) 4 A
  • (C) 1 A
  • (D) 8 A
Show Answer

Answer: A – Each parallel branch sees the full battery voltage, so I1 = V / R1 = 12 / 6 = 2 A.

Q41. In a series circuit, a 12 V battery is connected to a 4 Ω resistor in series with a 12 Ω resistor, and a capacitor is connected in parallel with the 12 Ω resistor. In steady state, what is the potential difference across the capacitor?

  • (A) 18 V
  • (B) 4.5 V
  • (C) 9 V
  • (D) 36 V
Show Answer

Answer: C – In steady state no current flows through the capacitor branch, so the capacitor voltage equals the voltage across the 12 Ω resistor: VC = V R2 / (R1 + R2) = 12 × 12 / 16 = 9 V.

Unit 12: Magnetic Fields and Electromagnetism

Q42. A particle of charge 4 μC moves at 2 × 104 m/s perpendicular to a uniform magnetic field of 0.5 T. What is the magnitude of the magnetic force on the particle?

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

Answer: A – F = q v B sin 90° = (4 × 10-6)(2 × 104)(0.5) = 40 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 0.5 T. What is the radius of its circular path?

  • (A) 48 cm
  • (B) 24 cm
  • (C) 12 cm
  • (D) 96 cm
Show Answer

Answer: B – r = m v / (q B) = (6.4 × 10-27)(6 × 106) / ((3.2 × 10-19)(0.5)) = 24 cm.

Q44. A straight wire of length 0.5 m carries a current of 5 A perpendicular to a uniform magnetic field of 50 mT. What is the magnitude of the magnetic force on the wire?

  • (A) 250 mN
  • (B) 62.5 mN
  • (C) 125 mN
  • (D) 500 mN
Show Answer

Answer: C – F = I L B = (5)(0.5)(50 × 10-3) = 125 mN.

Q45. A long straight wire carries a current of 2 A. What is the magnitude of the magnetic field at a distance of 20 cm from the wire?

  • (A) 4 μT
  • (B) 2 μT
  • (C) 1 μT
  • (D) 8 μT
Show Answer

Answer: B – B = μ0 I / (2π r) = (4π × 10-7)(2) / (2π × 0.2) = 2 μT.

Q46. A circular loop of radius 5 cm carries a current of 4 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 101 μT
  • (B) 25.1 μT
  • (C) 201 μT
  • (D) 50.3 μT
Show Answer

Answer: D – B = μ0 I / (2R) = (4π × 10-7)(4) / (2 × 0.05) = 50.3 μT.

Q47. A long solenoid has 1000 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) 7.54 mT
  • (B) 1.88 mT
  • (C) 3.77 mT
  • (D) 15.1 mT
Show Answer

Answer: C – B = μ0 n I = (4π × 10-7)(1000)(3) = 3.77 mT.

Q48. Two long parallel wires 10 cm apart carry currents of 4 A and 2 A in opposite directions. What is the magnitude of the net magnetic field at the midpoint between the wires?

  • (A) 48 μT
  • (B) 12 μT
  • (C) 24 μT
  • (D) 96 μT
Show Answer

Answer: C – At the midpoint each wire contributes B = μ0 I / (2π r) with r = 0.05 m, and for opposite currents the fields point the same way, so Bnet = μ0(I1 + I2) / (2π r) = 24 μT.

Q49. A rectangular coil of 10 turns and area 0.04 m2 carries a current of 2 A in a uniform magnetic field of 0.1 T. When the plane of the coil is parallel to the field, what is the magnitude of the torque on the coil?

  • (A) 160 mN·m
  • (B) 40 mN·m
  • (C) 80 mN·m
  • (D) 320 mN·m
Show Answer

Answer: C – Maximum torque τ = N I A B = (10)(2)(0.04)(0.1) = 80 mN·m.

Q50. In a velocity selector, a charged particle passes undeflected through crossed electric and magnetic fields of 2000 N/C and 40 mT. What is the speed of the particle?

  • (A) 1.00 × 105 m/s
  • (B) 5.00 × 104 m/s
  • (C) 2.50 × 104 m/s
  • (D) 2.00 × 105 m/s
Show Answer

Answer: B – For no deflection qE = qvB, so v = E / B = 2000 / (40 × 10-3) = 5.00 × 104 m/s.

Unit 13: Electromagnetic Induction

Q51. A flat loop of area 0.04 m2 is placed in a uniform magnetic field of 200 mT. The angle between the field and the normal to the loop is 60°. What is the magnetic flux through the loop?

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

Answer: A – Φ = B A cos θ = (200 × 10-3)(0.04) cos 60° = 4 mWb.

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

  • (A) 200 mV
  • (B) 50 mV
  • (C) 400 mV
  • (D) 100 mV
Show Answer

Answer: D – |ε| = A ΔB / Δt = (0.1)(0.2) / (0.2) = 100 mV.

Q53. A conducting rod of length 1 m moves at 3 m/s perpendicular to a uniform magnetic field of 0.5 T, with the rod perpendicular to both its velocity and the field. What is the motional emf between the ends of the rod?

  • (A) 3 V
  • (B) 0.75 V
  • (C) 6 V
  • (D) 1.5 V
Show Answer

Answer: D – ε = B L v = (0.5)(1)(3) = 1.5 V.

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

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

Answer: D – Increasing into-page flux requires an opposing out-of-page induced field, which corresponds to a counterclockwise current.

Q55. A solenoid has 200 turns, a cross-sectional area of 40 cm2, and a length of 0.4 m. What is its self-inductance?

  • (A) 1.01 mH
  • (B) 0.503 mH
  • (C) 0.251 mH
  • (D) 2.01 mH
Show Answer

Answer: B – L = μ0 N2 A / l = (4π × 10-7)(200)2(40 × 10-4) / (0.4) = 0.503 mH.

Q56. An inductor of 50 mH is connected in series with a 20 Ω resistor. What is the time constant of this LR circuit?

  • (A) 5 ms
  • (B) 1.25 ms
  • (C) 10 ms
  • (D) 2.5 ms
Show Answer

Answer: D – τ = L / R = (50 × 10-3) / 20 = 2.5 ms.

Q57. A series LR circuit with battery voltage 12 V and resistance 6 Ω is switched on at t = 0. After a time equal to 2 time constant(s), what is the current in the circuit?

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

Answer: A – I(t) = (V/R)(1 – e-t/τ) = (12/6)(1 – e-2) = 1.73 A.

Q58. An LC circuit consists of an inductor of 20 mH and a capacitor of 40 μF. What is its natural frequency of oscillation?

  • (A) 356 Hz
  • (B) 89 Hz
  • (C) 712 Hz
  • (D) 178 Hz
Show Answer

Answer: D – f = 1 / (2π√(LC)) = 1 / (2π√((20 × 10-3)(40 × 10-6))) = 178 Hz.

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

  • (A) 0.632 A
  • (B) 0.158 A
  • (C) 1.26 A
  • (D) 0.316 A
Show Answer

Answer: D – Energy conservation gives (1/2) C V2 = (1/2) L Imax2, so Imax = V √(C/L) = 20 √((10 × 10-6) / (40 × 10-3)) = 0.316 A.

Q60. A coil of 20 turns and area 0.02 m2 rotates in a uniform magnetic field of 0.2 T with angular speed 20 rad/s. What is the maximum induced emf in the coil?

  • (A) 3.2 V
  • (B) 0.8 V
  • (C) 6.4 V
  • (D) 1.6 V
Show Answer

Answer: D – εmax = N B A ω = (20)(0.2)(0.02)(20) = 1.6 V.

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