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Class 12 · CBSE

CBSE Class 12 Physics Part 1 Exercise Solutions

CBSE Class 12 Physics Part 1 NCERT exercise solutions with answer patterns, key concepts, and exam tips.

Updated 15 June 202631 min read

How to Use This Page

This page provides CBSE Class 12 Physics Part 1 exercise solutions structured for board exam preparation. Every section mirrors the question patterns you will face in the examination — MCQs, short answers, long answers, and numericals.

Before you begin, read the relevant NCERT chapter thoroughly. Then attempt the exercise questions on your own. Only after writing your own answer should you compare it with the solutions provided here. Pay close attention to the keywords, diagrams, and answer structure — these are what the CBSE marksheet rewards.

For numerical problems, always write the formula first, substitute values with correct units, and state the final answer with its SI unit. Skipping units or intermediate steps costs marks even if the final number is correct. Practise timed sessions: aim to solve a 2-mark question in 3 minutes and a 5-mark question in 8-10 minutes.

Use the Key Concepts section as a quick revision tool before tests. The Common Mistakes and Exam Checklist sections are designed to catch the errors that repeatedly appear in board examiner reports.

Key Concepts (Quick Recap)

Chapter 1: Electric Charges and Fields

  • Electric charge is an intrinsic property of matter; SI unit is coulomb (C). Charge is quantised: q = ne, where e = 1.6 x 10^-19 C.
  • Coulomb's Law: F = (1 / 4piepsilon0) x (q1 q2) / r^2, where epsilon0 = 8.854 x 10^-12 C^2 N^-1 m^-2 and 1/4piepsilon0 = 9 x 10^9 N m^2 C^-2.
  • Superposition principle: Net force on a charge is the vector sum of individual forces from all other charges.
  • Electric field E at a point = force per unit positive test charge: E = F/q0. SI unit: N C^-1 or V m^-1.
  • Electric field lines start on positive charges and end on negative charges; they never cross.
  • Electric dipole: Two equal and opposite charges (+q, -q) separated by distance 2a. Dipole moment p = q x 2a (from -q to +q). SI unit: C m.
  • Dipole field: On axial line, E = (1/4piepsilon0) x 2p/r^3; on equatorial line, E = (1/4piepsilon0) x p/r^3 (both for r >> a).
  • Torque on dipole in uniform field: tau = p x E, magnitude = pE sin theta.
  • Gauss's Theorem: Total electric flux through a closed surface = (1/epsilon0) x charge enclosed. phi = oint E . dA = q_enc / epsilon0.
  • Applications of Gauss's Law:
  • Infinite line charge: E = lambda / (2 pi epsilon0 r)
  • Infinite plane sheet: E = sigma / (2 epsilon0)
  • Conducting sphere (outside): E = (1/4piepsilon0) x Q/r^2; (inside): E = 0

Chapter 2: Electrostatic Potential and Capacitance

  • Electric potential V at a point = work done per unit charge in bringing a test charge from infinity to that point. V = (1/4piepsilon0) x q/r.
  • Potential difference: V_A - V_B = W_{AB}/q. SI unit: volt (V).
  • Relation between E and V: E = -dV/dr (electric field is in direction of decreasing potential).
  • Potential due to dipole: On axial line, V = (1/4piepsilon0) x p cos theta / r^2; on equatorial line, V = 0.
  • Equipotential surfaces: Surfaces where V is constant. E is perpendicular to equipotential surfaces; no work is done moving a charge along an equipotential surface.
  • Work done in rotating a dipole from theta1 to theta2: W = pE(cos theta1 - cos theta2). Work to rotate from 0 to 180 degree = 2pE.
  • Electrostatic potential energy of a system of charges: U = (1/4piepsilon0) x sum(q_i q_j / r_ij).
  • Capacitance: C = Q/V. SI unit: farad (F). For a parallel plate capacitor: C = epsilon0 A / d.
  • Dielectrics: With dielectric of dielectric constant K, C' = K x C. Electric field inside dielectric: E' = E/K.
  • Capacitors in series: 1/C_s = 1/C1 + 1/C2 + ... ; in parallel: C_p = C1 + C2 + ...
  • Energy stored in a capacitor: U = (1/2) CV^2 = (1/2) QV = Q^2/(2C).
  • Van de Graaff generator: Uses electrostatic charging to build up very high potentials on a spherical shell.

Chapter 3: Current Electricity

  • Electric current: I = dQ/dt. SI unit: ampere (A). Conventional current flows from higher to lower potential.
  • Ohm's Law: V = IR. Resistance R = rho l / A, where rho is resistivity.
  • Drift velocity: v_d = (eE / m) x tau = I / (n A e), where tau is relaxation time, n is number density.
  • Mobility: mu = |v_d| / E = e tau / m. Current density j = sigma E = n e v_d.
  • Kirchhoff's Laws: (1) Junction rule — sum of currents at a junction = 0. (2) Loop rule — algebraic sum of potential differences around a closed loop = 0.
  • Wheatstone bridge: When balanced, P/Q = R/S, and no current flows through the galvanometer.
  • Metre bridge: Based on Wheatstone bridge. R/S = l/(100 - l). Used to find unknown resistance.
  • Potentiometer: Measures emf without drawing current. emf balancing: emf1/emf2 = l1/l2. Internal resistance: r = R(l1 - l2)/l2.
  • Cells in series: Total emf = emf1 + emf2 + ... ; total internal resistance = r1 + r2 + ...
  • Cells in parallel (identical): Total emf = emf; effective internal resistance = r/n for n cells.
  • Temperature dependence: R_T = R_0[1 + alpha(T - T_0)], where alpha is the temperature coefficient of resistance.

Chapter 4: Moving Charges and Magnetism

  • Magnetic force on a moving charge: F = q(v x B). Magnitude: F = qvB sin theta. Direction by Fleming's Left Hand Rule.
  • Biot-Savart Law: dB = (mu0 / 4pi) x (I dl x r_hat) / r^2. mu0 = 4pi x 10^-7 T m A^-1.
  • Magnetic field on axis of circular loop (radius R, distance x): B = (mu0 I R^2) / [2(R^2 + x^2)^(3/2)]. At centre: B = mu0 I / (2R).
  • Ampere's Circuital Law: oint B . dl = mu0 x I_enclosed.
  • Applications: Infinite straight wire: B = mu0 I / (2pi r). Solenoid: B = mu0 n I. Toroid: B = mu0 n I / (2pi r).
  • Force between two parallel conductors: F/l = (mu0 I1 I2) / (2pi d). Definition of ampere uses this.
  • Torque on current loop in magnetic field: tau = n I A B sin theta = m B sin theta, where m = nIA is magnetic moment.
  • Moving coil galvanometer: I = (k / nBA) x theta. Current sensitivity = theta/I = nBA/k. Converted to ammeter with shunt S = (I_g R_g) / (I - I_g). Converted to voltmeter with series resistance R = V/I_g - R_g.

Chapter 5: Magnetism and Matter

  • Magnetic dipole moment of a current loop: m = nIA. Direction by right-hand rule.
  • Bar magnet as solenoid: m = NIA. Field on axial line: B = (mu0 / 4pi) x 2m / r^3. On equatorial line: B = (mu0 / 4pi) x m / r^3.
  • Magnetic field lines of a bar magnet: continuous closed loops from N to S outside, S to N inside.
  • Earth's magnetism: Components — B_H = B cos delta (horizontal), B_V = B sin delta (vertical), where delta is angle of dip. tan delta = B_V / B_H.
  • Magnetic elements: Declination (angle between geographic and magnetic meridians), Dip/Inclination (angle with horizontal), Horizontal component B_H.
  • Classification of magnetic materials:
  • Diamagnetic: Weakly repelled, chi is small and negative, mu_r < 1 (e.g., Cu, Bi, Au).
  • Paramagnetic: Weakly attracted, chi is small and positive, mu_r > 1 (e.g., Al, Na, Ca).
  • Ferromagnetic: Strongly attracted, chi is very large and positive, mu_r >> 1 (e.g., Fe, Co, Ni). Above Curie temperature, become paramagnetic.
  • Hysteresis: Lag of magnetisation behind the magnetising field. Area of hysteresis loop = energy loss per cycle. Retentivity and coercivity are key parameters.
  • Permanent magnets need high retentivity and high coercivity (hard ferromagnetic materials). Electromagnets need low coercivity (soft ferromagnetic materials).

Chapter 6: Electromagnetic Induction

  • Faraday's Law: The induced emf in a circuit equals the negative rate of change of magnetic flux through the circuit. emf = -d(phi)/dt, where phi = B . A = BA cos theta.
  • Lenz's Law: The direction of induced current opposes the change in magnetic flux that produces it. This is a consequence of conservation of energy.
  • Motional emf: emf = Blv (conductor of length l moving with velocity v perpendicular to B).
  • Self-inductance: emf = -L (dI/dt). L = phi/I. For a solenoid: L = mu0 n^2 A l. SI unit: henry (H).
  • Mutual inductance: emf = -M (dI2/dt) in coil 1 due to current change in coil 2. M = mu0 n1 n2 A l (for coaxial solenoids).
  • Eddy currents: Circulating currents induced in bulk conductors by changing magnetic flux. Used in induction furnaces, electromagnetic brakes, and dead-beat galvanometers. Undesired eddy currents cause energy loss in transformers (minimized by laminated core).
  • Energy stored in an inductor: U = (1/2) LI^2. This energy is stored in the magnetic field: u = B^2 / (2 mu0).
  • AC generator: emf = NBA omega sin(omega t) = emf_0 sin(omega t), where emf_0 = NBA omega is the peak emf.

Chapter 7: Alternating Current

  • AC voltage and current: V = V_0 sin(omega t), I = I_0 sin(omega t - phi), where omega = 2pi f.
  • RMS value: I_rms = I_0 / sqrt(2) = 0.707 I_0. V_rms = V_0 / sqrt(2).
  • Pure resistor: V and I in phase. Power dissipated: P = I_rms^2 R = V_rms I_rms.
  • Pure inductor: V leads I by 90 degree. Inductive reactance: X_L = omega L = 2 pi f L.
  • Pure capacitor: I leads V by 90 degree. Capacitive reactance: X_C = 1 / (omega C) = 1 / (2 pi f C).
  • Series LCR circuit: Impedance Z = sqrt(R^2 + (X_L - X_C)^2). Phase angle: tan phi = (X_L - X_C) / R.
  • Resonance: When X_L = X_C, Z = R (minimum), current is maximum. Resonant frequency: f_r = 1 / (2 pi sqrt(LC)).
  • Quality factor (Q-factor): Q = (1/R) sqrt(L/C) = omega_r L / R = 1 / (omega_r C R). Higher Q means sharper resonance.
  • Power in AC circuit: P = V_rms I_rms cos phi. cos phi is the power factor = R / Z.
  • For pure R: cos phi = 1, P = V_rms I_rms.
  • For pure L or pure C: cos phi = 0, P = 0 (wattless current).
  • LC oscillations: Charge oscillates as q = q_0 cos(omega t), omega = 1/sqrt(LC). Analogous to SHM.
  • Transformer: Works on mutual induction. V_s / V_p = N_s / N_p = n (turns ratio). I_s / I_p = N_p / N_s. Efficiency = (P_output / P_input) x 100%.
  • Energy losses in transformers: Copper loss (I^2 R), Eddy current loss (laminated core reduces this), Hysteresis loss, Flux leakage, Humming loss.

I. Choose the Best Answer (MCQs)

1. A point charge +q is placed at the centre of a spherical shell. What is the electric flux through the shell surface if another charge +2q is placed outside the shell? (a) q/epsilon0 (b) 2q/epsilon0 (c) 3q/epsilon0 (d) zero
Answer(a) — By Gauss's law, flux depends only on the charge enclosed. The charge outside the shell does not contribute to the net flux through the closed surface. Enclosed charge = q, so phi = q/epsilon0.
2. Two capacitors of 4 microfarad and 6 microfarad are connected in series across a 100 V supply. The charge on each capacitor is: (a) 400 microC (b) 600 microC (c) 240 microC (d) 1000 microC
Answer(c) — Equivalent capacitance in series: C_eq = (4 x 6)/(4 + 6) = 24/10 = 2.4 microfarad. Q = C_eq x V = 2.4 x 100 = 240 microC. In series, each capacitor stores the same charge.
3. A copper wire of resistance R is stretched to double its length without changing its volume. Its new resistance will be: (a) R (b) 2R (c) 4R (d) R/4
Answer(c) — Volume = A x l is constant. If l becomes 2l, then A becomes A/2. New resistance = rho(2l)/(A/2) = 4 x rho l/A = 4R.
4. The magnetic field at the centre of a circular coil of radius R carrying current I is B. If the radius is doubled and the current is halved, the magnetic field at the centre becomes: (a) B/2 (b) B (c) 2B (d) B/4
Answer(d) — B = mu0 I / (2R). New B' = mu0 (I/2) / (2 x 2R) = (1/4) x mu0 I / (2R) = B/4.
5. A metallic rod of length l is rotated about one end with angular velocity omega in a uniform magnetic field B perpendicular to the plane of rotation. The emf induced between its ends is: (a) Blv (b) (1/2) B omega l^2 (c) B omega l^2 (d) 2 B omega l^2
Answer(b) — The area swept per second = (1/2) omega l^2. Rate of change of flux = B x (1/2) omega l^2. Hence emf = (1/2) B omega l^2.
6. In a series LCR circuit at resonance: (a) impedance is maximum and current is minimum (b) impedance is maximum and current is maximum (c) impedance is minimum and equal to R, and current is maximum (d) impedance is minimum and equal to zero
Answer(c) — At resonance, X_L = X_C, so Z = sqrt(R^2 + 0) = R (minimum). Since Z is minimum, current I = V/Z is maximum.
7. The power factor of a series LCR circuit at resonance is: (a) zero (b) 0.5 (c) 0.707 (d) 1
Answer(d) — At resonance, the circuit is purely resistive (Z = R). Phase angle phi = 0, so cos phi = 1. Power factor = R/Z = R/R = 1.
8. A charged particle enters a uniform magnetic field perpendicular to its direction of motion. Its: (a) energy changes but momentum does not (b) momentum changes but energy does not (c) both energy and momentum change (d) neither energy nor momentum change
Answer(b) — The magnetic force is always perpendicular to velocity, so no work is done; kinetic energy remains constant. However, the direction of velocity changes, so momentum (a vector) changes in direction.
9. In a Wheatstone bridge, the galvanometer shows null deflection when P = 10 ohm, Q = 30 ohm, and R = 40 ohm. The value of S is: (a) 120 ohm (b) 80 ohm (c) 60 ohm (d) 20 ohm
Answer(a) — At balance: P/Q = R/S, so S = (Q x R)/P = (30 x 40)/10 = 120 ohm.
10. A transformer with 200 turns in the primary and 1000 turns in the secondary is connected to a 220 V AC source. The secondary voltage is: (a) 44 V (b) 220 V (c) 1100 V (d) 440 V
Answer(c) — V_s/V_p = N_s/N_p. V_s = V_p x (N_s/N_p) = 220 x (1000/200) = 1100 V. This is a step-up transformer.

II. Fill in the Blanks

1. The SI unit of electric field intensity is N C^-1 (or V m^-1). 2. The relation between electric field and electric potential is given by E = -dV/dr. 3. In a series combination of resistors, the current through each resistor is the same. 4. The magnetic force on a stationary charge placed in a magnetic field is zero. 5. The Biot-Savart law gives the magnetic field due to a current element. 6. The phenomenon of producing induced emf in a coil due to change in current in the same coil is called self-induction. 7. At resonance in a series LCR circuit, the impedance is equal to R. 8. The power factor of a purely inductive AC circuit is zero. 9. In a step-down transformer, the number of turns in the secondary is less than the number of turns in the primary. 10. The energy stored in a capacitor of capacitance C charged to voltage V is given by (1/2) CV^2.

III. True or False

1. Electric field lines can form closed loops.
AnswerFalse — Correction: Electric field lines start on positive charges and end on negative charges. They do not form closed loops. (Magnetic field lines form closed loops.)
2. The electric potential at a point midway between two equal positive charges is zero.
AnswerFalse — Correction: The potential at the midpoint is twice the potential due to a single charge (V = 2kq/r). It is the electric field that is zero at the midpoint between two equal like charges, not the potential.
3. A potentiometer can measure emf without drawing any current from the cell.
AnswerTrue — At balance, no current flows through the galvanometer, so the measured emf equals the true emf. This is why a potentiometer is preferred over a voltmeter for emf measurement.
4. The force between two parallel current-carrying conductors is attractive if the currents flow in opposite directions.
AnswerFalse — Correction: Parallel conductors with currents in the same direction attract, and currents in opposite directions repel. Like currents attract, unlike currents repel.
5. A diamagnetic material has a relative permeability slightly less than 1.
AnswerTrue — Diamagnetic materials are weakly repelled by a magnetic field. Their magnetic susceptibility is small and negative, so mu_r = 1 + chi is slightly less than 1.
6. Lenz's law is a consequence of the law of conservation of energy.
AnswerTrue — If the induced current aided the change in flux, it would produce energy from nothing. Lenz's law ensures that work must be done against the opposing force, consistent with energy conservation.
7. In a pure capacitive circuit, the power consumed is infinite.
AnswerFalse — Correction: In a pure capacitive circuit, the phase angle between voltage and current is 90 degree, so cos phi = 0 and power consumed P = V_rms I_rms cos(90 degree) = 0. The current is called wattless current.
8. The Q-factor of a resonant circuit is a measure of the sharpness of resonance.
AnswerTrue — Higher Q-factor means a narrower bandwidth and sharper resonance peak. Q = omega_r L / R = (1/R) sqrt(L/C).

IV. Match the Following

Column AColumn B
1. Gauss's Law(e) phi = q_enc / epsilon0
2. Capacitance(g) C = Q / V
3. Ohm's Law(j) V = IR
4. Biot-Savart Law(a) dB = (mu0 / 4pi) x (I dl sin theta) / r^2
5. Faraday's Law(c) emf = -d(phi)/dt
6. Self-inductance(i) emf = -L (dI/dt)
7. Resonance condition(h) X_L = X_C
8. Power factor(b) cos phi = R / Z
9. Energy in capacitor(f) U = (1/2) CV^2
10. Transformer equation(d) V_s / V_p = N_s / N_p

V. Answer Briefly

Q1. Why must electrostatic field lines be normal to the surface of a conductor at every point on the surface?
Answer

If the electric field had a tangential component at the surface of a conductor, free electrons would experience a force along the surface and would move, creating a surface current. In electrostatic equilibrium, there can be no current inside the conductor. Hence, the electric field must be entirely normal (perpendicular) to the surface. This also means the surface of a conductor is an equipotential surface.

Q2. Derive the expression for the equivalent capacitance of two capacitors connected in parallel.
Answer

When capacitors C1 and C2 are connected in parallel across the same potential difference V, the charge on each is Q1 = C1V and Q2 = C2V. The total charge Q = Q1 + Q2 = (C1 + C2)V. Since Q = C_eq V, comparing we get C_eq = C1 + C2. For n capacitors in parallel: C_eq = C1 + C2 + C3 + ... + Cn. The equivalent capacitance in parallel is the sum of individual capacitances — it is always greater than the largest individual capacitance.

Q3. State the principle of a potentiometer. Why is it preferred over a voltmeter for measuring emf?
Answer

Principle: When a constant current flows through a uniform wire of length l, the potential drop across any portion of the wire is directly proportional to the length of that portion: V = kl, where k is the potential gradient. A potentiometer is preferred over a voltmeter because at the balance point, no current is drawn from the cell being measured. A voltmeter always draws some current, so it reads the terminal voltage (emf - Ir) rather than the true emf. The potentiometer uses the null-deflection method, giving an accurate measurement.

Q4. Derive the expression for the magnetic field at the centre of a circular coil of radius R carrying current I.
Answer

Using Biot-Savart Law: dB = (mu0 / 4pi) x (I dl sin theta) / r^2. For a point at the centre, r = R and theta = 90 degree, so sin theta = 1. dB = (mu0 / 4pi) x (I dl) / R^2. Integrating over the entire loop of circumference 2piR: B = (mu0 I / 4pi R^2) x oint dl = (mu0 I / 4pi R^2) x 2piR = mu0 I / (2R). For a coil of n turns: B = mu0 n I / (2R).

Q5. Define magnetic declination and angle of dip.
Answer

Magnetic declination is the angle between the geographic meridian and the magnetic meridian at a given place on the Earth's surface. It indicates how far the magnetic north deviates from true (geographic) north. Angle of dip (inclination) is the angle made by the Earth's total magnetic field with the horizontal direction at a given place. At the magnetic equator, dip is 0 degree; at the magnetic poles, dip is 90 degree. Components: B_H = B cos delta, B_V = B sin delta, where delta is the angle of dip.

Q6. State Lenz's law. How does it illustrate the law of conservation of energy?
Answer

Lenz's law states that the direction of the induced emf (and hence the induced current) is such that it opposes the change in magnetic flux that produces it. The negative sign in Faraday's law (emf = -dphi/dt) represents this opposing nature. This law illustrates conservation of energy because if the induced current aided the change in flux, it would generate energy without any external work — a clear violation. The opposing force means that mechanical work must be done against this opposition, and this work appears as electrical energy in the circuit. Thus, energy is conserved.

Q7. What is the power factor of an AC circuit? What is its value for a purely resistive, purely inductive, and purely capacitive circuit?
Answer

The power factor of an AC circuit is cos phi, where phi is the phase angle between the voltage and the current. It determines what fraction of the apparent power (V_rms x I_rms) is the true power consumed: P = V_rms I_rms cos phi. - Purely resistive circuit: phi = 0, cos phi = 1. All power is consumed. - Purely inductive circuit: phi = 90 degree, cos phi = 0. No power consumed (wattless current). - Purely capacitive circuit: phi = 90 degree, cos phi = 0. No power consumed (wattless current).

VI. Answer in Detail

Q1. Derive the expression for the electric field due to an infinite plane sheet of charge using Gauss's theorem.
Answer

Consider an infinite plane sheet with uniform surface charge density sigma (C m^-2). By symmetry, the electric field is perpendicular to the sheet and points away from it (for sigma > 0) on both sides. Choose a Gaussian surface in the form of a cylinder (pillbox) of cross-sectional area A, with its flat faces parallel to the sheet on either side. The curved surface is parallel to E, so no flux passes through it. The flux through each flat face is E x A. Total flux: phi = E x A + E x A = 2EA. Charge enclosed: q_enc = sigma x A. By Gauss's Law: 2EA = sigma A / epsilon0. Therefore: E = sigma / (2 epsilon0). Key observations: (1) The field is independent of the distance from the sheet — it is uniform. (2) The field direction reverses on the two sides of the sheet. (3) This result holds for any infinite plane with uniform charge density, whether it is a conducting or non-conducting sheet. For a conducting sheet, since charge resides on both surfaces, E = sigma / epsilon0 (twice the non-conducting value) outside the conductor.

Q2. Derive the condition for balance of a Wheatstone bridge. How is it used in a metre bridge to determine an unknown resistance?
Answer

Wheatstone Bridge Balance Condition: A Wheatstone bridge consists of four resistors P, Q, R, S arranged in a diamond with a galvanometer G between the middle points. A battery is connected across one diagonal. When no current flows through the galvanometer (null deflection), the bridge is balanced. At balance, the potential at point B equals the potential at point D. So: V_A - V_B = I_1 P and V_A - V_D = I_2 R. Since V_B = V_D: I_1 P = I_2 R ... (1) Similarly, V_B - V_C = I_1 Q and V_D - V_C = I_2 S. So: I_1 Q = I_2 S ... (2) Dividing (1) by (2): P/Q = R/S — this is the balance condition. Metre Bridge: A metre bridge is a practical form of the Wheatstone bridge. A uniform wire AC of length 1 m (100 cm) is mounted on a wooden board with a metre scale. A jockey can slide along the wire to find the null point. P and Q are replaced by lengths l and (100 - l) of the wire. Since the wire is uniform, resistance is proportional to length: P = rho l/A and Q = rho(100 - l)/A. Balance condition becomes: R/S = l/(100 - l). To find unknown resistance S: Connect known resistance R in the left gap and unknown S in the right gap. Slide the jockey to find null point at length l. Then S = R x (100 - l) / l. Precautions: (1) Ensure all connections are clean and tight. (2) The null point should be near the middle of the wire (between 30 cm and 70 cm) for minimum error. (3) Never press the jockey hard — it can damage the wire.

Q3. A parallel plate capacitor of capacitance 5 microfarad is connected to a 200 V supply. A dielectric slab of dielectric constant 4 is inserted between the plates, fully occupying the space. Calculate (a) the charge on the capacitor before the dielectric, (b) the charge and potential after the dielectric is inserted with the battery still connected, (c) the charge and potential if the battery is disconnected before the dielectric is inserted.
GivenC = 5 microF = 5 x 10^-6 F, V = 200 V, K = 4. (a) Charge before dielectric: Q = CV = 5 x 10^-6 x 200 = 1000 microC = 1.0 x 10^-3 C (b) Battery remains connected (voltage stays constant): When the dielectric is inserted, the capacitance becomes: C' = K x C = 4 x 5 = 20 microF Since the battery is still connected, V stays at 200 V. Q' = C' x V = 20 x 10^-6 x 200 = 4000 microC = 4.0 x 10^-3 C Potential remains 200 V. (c) Battery disconnected first (charge stays constant): Charge remains Q = 1000 microC = 1.0 x 10^-3 C. New capacitance: C' = K x C = 20 microF. New potential: V' = Q / C' = (1.0 x 10^-3) / (20 x 10^-6) = 50 V Summary: When the battery stays connected, V is constant and Q increases by factor K. When the battery is disconnected, Q is constant and V decreases by factor K.
Q4. Derive the expression for the force per unit length between two parallel current-carrying conductors. Define the ampere on this basis.
Answer

Consider two infinitely long parallel conductors separated by distance d, carrying currents I1 and I2 in the same direction. Step 1: Magnetic field at the location of conductor 2 due to conductor 1: B1 = mu0 I1 / (2 pi d) — directed perpendicular to the plane containing the conductors. Step 2: Force on a length l of conductor 2 due to this field: F2 = I2 x l x B1 = I2 x l x (mu0 I1) / (2 pi d) By Newton's Third Law, an equal and opposite force acts on conductor 1. Force per unit length: F/l = mu0 I1 I2 / (2 pi d) Direction: If currents are in the same direction, the force is attractive (conductors pull each other). If currents are in opposite directions, the force is repulsive. Definition of ampere: When I1 = I2 = 1 A and d = 1 m: F/l = (4 pi x 10^-7 x 1 x 1) / (2 pi x 1) = 2 x 10^-7 N m^-1 One ampere is that constant current which, when maintained in each of two infinitely long straight parallel conductors of negligible circular cross-section placed 1 metre apart in vacuum, produces between them a force equal to 2 x 10^-7 newton per metre of length.

Q5. With the help of a labelled diagram, explain the principle, construction, and working of an AC generator.
Answer

Principle: An AC generator works on the phenomenon of electromagnetic induction. When a coil rotates in a uniform magnetic field, the magnetic flux through the coil changes continuously, inducing an emf in the coil (Faraday's law). Construction: An AC generator consists of: (1) Armature: A rectangular coil of many turns of insulated copper wire wound on a soft iron core. (2) Field magnet: A strong permanent magnet or electromagnet that provides a uniform radial magnetic field. (3) Slip rings: Two smooth circular rings attached to the two ends of the armature coil. They rotate with the coil and maintain continuous contact with the brushes. (4) Brushes: Two carbon or metallic brushes that press against the slip rings to collect the induced current and deliver it to the external circuit. Working: When the armature coil of N turns and area A rotates with angular velocity omega in a magnetic field B, the flux through the coil at time t is: phi = NBA cos(omega t) By Faraday's law: emf = -d(phi)/dt = NBA omega sin(omega t) So emf = emf_0 sin(omega t), where emf_0 = NBA omega is the peak emf. The current varies sinusoidally: I = I_0 sin(omega t). Key features: - When omega t = 0 or 180 degree, the coil is perpendicular to B and emf = 0 (maximum flux, minimum change). - When omega t = 90 degree or 270 degree, the coil is parallel to B and emf is maximum (minimum flux, maximum rate of change). - The direction of current reverses every half rotation, producing alternating current.

Q6. A series LCR circuit with R = 40 ohm, L = 5.0 H, and C = 80 microfarad is connected to a 230 V, 50 Hz supply. Calculate (a) the impedance, (b) the current, (c) the phase angle, (d) the power factor, and (e) the power consumed. Determine the resonant frequency.
GivenR = 40 ohm, L = 5.0 H, C = 80 x 10^-6 F, V_rms = 230 V, f = 50 Hz. Angular frequency: omega = 2 pi f = 2 x 3.14 x 50 = 314 rad/s Inductive reactance: X_L = omega L = 314 x 5.0 = 1570 ohm Capacitive reactance: X_C = 1/(omega C) = 1/(314 x 80 x 10^-6) = 1/0.02512 = 39.8 ohm Net reactance: X_L - X_C = 1570 - 39.8 = 1530.2 ohm (a) Impedance: Z = sqrt(R^2 + (X_L - X_C)^2) = sqrt(40^2 + 1530.2^2) = sqrt(1600 + 2341512) = sqrt(2343112) = 1530.7 ohm (approximately) (b) Current: I_rms = V_rms / Z = 230 / 1530.7 = 0.150 A (approximately) (c) Phase angle: tan phi = (X_L - X_C) / R = 1530.2 / 40 = 38.26 phi = arctan(38.26) = 88.5 degree (approximately) The voltage leads the current by 88.5 degree (since X_L > X_C, the circuit is inductive). (d) Power factor: cos phi = R / Z = 40 / 1530.7 = 0.0261 (e) Power consumed: P = V_rms I_rms cos phi = 230 x 0.150 x 0.0261 = 0.90 W (approximately) Alternatively: P = I_rms^2 x R = (0.150)^2 x 40 = 0.90 W Resonant frequency: f_r = 1 / (2 pi sqrt(LC)) = 1 / (2 x 3.14 x sqrt(5.0 x 80 x 10^-6)) = 1 / (6.28 x sqrt(4 x 10^-4)) = 1 / (6.28 x 0.02) = 1 / 0.1256 = 7.96 Hz (approximately 8 Hz) Note: At such a low resonant frequency compared to 50 Hz, the circuit is highly inductive, explaining the large phase angle and very low power factor.

Common Mistakes to Avoid

1. Confusing electric field and electric potential: Electric field (E) is a vector quantity (N C^-1), while electric potential (V) is a scalar quantity (V). The relation E = -dV/dr is a differential relation — do not simply write E = V/r for non-uniform fields.

2. Incorrect series/parallel capacitance formulas: Students often swap the formulas. Remember: for capacitors, series adds like resistors in parallel (1/C_eq = 1/C1 + 1/C2) and parallel adds like resistors in series (C_eq = C1 + C2). Write the formula first, then substitute.

3. Sign errors in Kirchhoff's loop rule: When traversing a loop, assign signs consistently. EMF is positive when traversed from negative to positive terminal; potential drop across a resistor is negative when traversed in the direction of current. Pick a direction and stick with it.

4. Wrong direction for magnetic force: Use Fleming's Left Hand Rule correctly — Forefinger (Field), Middle finger (Current/velocity), Thumb (Force). Do not confuse with the right-hand rules used for field direction.

5. Forgetting the negative sign in Faraday's law: emf = -d(phi)/dt. The negative sign represents Lenz's law and is essential. Dropping it loses marks. Always state that the negative sign indicates opposing nature of induced emf.

6. Applying DC formulas to AC circuits: In AC circuits, always use impedance Z = sqrt(R^2 + (X_L - X_C)^2) instead of simple resistance. Voltage and current are out of phase — you cannot use V = IR directly. Use V = IZ and account for the phase angle. Also, remember that power P = V_rms I_rms cos phi, not simply V_rms I_rms.

Exam Checklist

  • [ ] Can I write Coulomb's law, Gauss's law, and derive electric field for point charge, infinite line, and infinite sheet?
  • [ ] Can I derive the capacitance of a parallel plate capacitor with and without a dielectric?
  • [ ] Can I apply Kirchhoff's laws to solve circuit problems and state the Wheatstone bridge condition?
  • [ ] Can I use the metre bridge and potentiometer formulas to solve numerical problems?
  • [ ] Can I derive the magnetic field on the axis of a circular loop and inside a solenoid using Biot-Savart and Ampere's law?
  • [ ] Can I calculate the force on a moving charge and a current-carrying conductor in a magnetic field, including torque on a dipole?
  • [ ] Can I state and apply Faraday's law and Lenz's law to motional emf and transformer problems?
  • [ ] Can I solve series LCR circuit problems for impedance, current, phase angle, resonance, and power factor?

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