Resistance R = V/I. From the graph pick any point (for example V = 2 V, I = 1 A), then R = 2/1 = 2 Ω. Thus the resistance is 2 Ω.
Total circumference = 2π(1) = 2π m. Each semicircle length = π m. Resistance per semicircle = (2 Ω/m)×π m = 2π Ω. Two equal resistances 2π Ω in parallel: R_eq = (2π × 2π)/(2π + 2π) = (4π^2)/(4π) = π Ω.
Power P = V^2/R = (240^2)/120 = 57600/120 = 480 W.
47 kΩ → digits 4 (yellow), 7 (violet), multiplier 10^3 (orange). Tolerance ±10% corresponds to silver. So Yellow–Violet–Orange–Silver.
Using the given band colours (as in the book figure) the decoded value is 100 kΩ. (Answer given in book: 100 kΩ.)
For equal length and same material, R ∝ 1/area ∝ 1/r^2. So R_A/R_B = (r_B/r_A)^2 = 3 ⇒ r_B/r_A = √3 ⇒ r_A/r_B = 1/√3.
If original resistance is R, cutting into two halves gives each R/2. Two halves in parallel give R_eq = (R/2)||(R/2) = R/4. Power P ∝ 1/R. So P2/P1 = R/R_eq = R/(R/4) = 4.
For same power P, R = V^2/P. So R_USA/R_India = (110^2)/(220^2) = (1/2)^2 = 1/4 ⇒ R_USA = R/4.
Total power = 15×40 + 5×100 + 5×80 + 1000 = 600 + 500 + 400 + 1000 = 2500 W. Current I = P/V = 2500/220 ≈ 11.36 A ⇒ choose 12 A (nearest standard fuse rating).
Total resistance = 9 V / 1 A = 9 Ω. Known resistances 2.5 + 3 = 5.5 Ω. So P = 9 − 5.5 = 3.5 Ω.
Three 15 Ω resistors in parallel give R_eq = 15/3 = 5 Ω. Current I = 5 V / 5 Ω = 1 A.
R_T = R_0[1 + α(T − T_0)]. Put R_T = 2, R_0 = 1 at T_0 = 20 °C and α = 0.00125: 2 = 1 + 0.00125(T − 20) ⇒ T − 20 = 800 ⇒ T = 820 °C.
E = I(R + r) ⇒ r = E/I − R = 2.1/0.2 − 10 = 10.5 − 10 = 0.5 Ω.
Metals (copper) have fewer phonon collisions at low T ⇒ resistance decreases. Semiconductor (germanium) has reduced carrier concentration at low T ⇒ resistance increases.
Joule heating H = I^2 R t. With R and t constant, H ∝ I^2, so plotting H vs I^2 yields a straight line through the origin.
Current I = dQ/dt; charge Q is a scalar. Although it has a direction of flow, current adds algebraically (signed) and does not follow vector addition, so it is treated as a scalar in circuit theory.
Current is a scalar because it represents the rate of charge flow (a scalar quantity) through a cross-section per unit time and direction is accounted for by sign convention; it does not obey vector addition rules.
Microscopically, current density J = nq v_d where n is charge carrier density, q charge per carrier and v_d drift velocity. For average current across area A, J = I/A.
Current density J is the current per unit cross-sectional area: J = I/A (vector quantity pointing in direction of charge flow).
Drift velocity depends on field strength and carrier scattering. Mobility μ = v_d/E (units m^2 V^{-1} s^{-1}) characterizes how easily carriers move in a material; higher μ means larger v_d for a given E.
Drift velocity v_d is the average velocity acquired by charge carriers under an electric field. Mobility μ is the proportionality constant between drift velocity and electric field: v_d = μE.
Using J = nq v_d and v_d = μE, we get J = nq μ E ≡ σ E. Inverse relation: E = ρ J with ρ = 1/σ.
Microscopic Ohm's law: J = σE, where J is current density, σ is conductivity (σ = nqμ) and E is electric field.
For a uniform conductor of length L and cross sectional area A, R = ρ L/A, and V = IR follows from E = V/L and J = σ E.
Macroscopic Ohm's law: V = IR, where V is potential difference across a conductor and I is the current through it; R is resistance (constant for ohmic conductors).
Examples: metallic resistor (ohmic over range), diode/filament lamp/semiconductor (non-ohmic).
Ohmic devices obey V ∝ I (constant R); their V–I graph is a straight line passing through origin. Non-ohmic devices do not follow a linear V–I relation; R may vary with V, I or temperature.
It quantifies how strongly a material opposes electric current. Conductivity σ = 1/ρ.
Resistivity ρ is the intrinsic property of a material defined by R = ρ L / A. Units: Ω·m.
Positive α means resistance increases with temperature (metals); negative α means resistance decreases with temperature (some semiconductors).
Temperature coefficient α is defined by α = (1/R)(dR/dT) at a given temperature; for small ΔT, R_T = R_0[1 + α (T − T_0)].
Key properties: zero resistivity → persistent currents, perfect diamagnetism, used in MRI, particle accelerators, lossless power transmission (in principle). Superconductivity disappears above T_c or beyond critical magnetic field/current.
Superconductors are materials that exhibit zero electrical resistance and expel magnetic fields (Meissner effect) below a critical temperature T_c.
Using circuit relations, P can be expressed as P = VI = I^2R = V^2/R. Energy in time t: E = ∫ P dt (for constant P, E = P t). Units: power in watts, energy in joules or kWh for billing.
Electric power P is the rate at which electrical energy is supplied or dissipated: P = VI. Electric energy E is energy consumed over time t: E = Pt.
A charge ΔQ moving through potential difference V gains energy ΔW = V ΔQ. Rate of energy transfer P = ΔW/Δt = V (ΔQ/Δt) = VI.
Power delivered P = (energy transferred per unit time) = (charge delivered per unit time) × (energy per unit charge) = I × V ⇒ P = VI.
Using Ohm's law V = IR, substitute into P = VI to get P = (IR)I = I^2 R, or P = V (V/R) = V^2 / R.
P = VI = I^2 R = V^2 / R.
This follows from charge conservation at the node.
Kirchhoff’s current rule (KCL): The algebraic sum of currents entering a junction equals the sum leaving it, or equivalently the sum of currents at a node is zero.
Follows from energy conservation: sum of emf and drops = 0 when following loop sign convention.
Kirchhoff’s voltage rule (KVL): The algebraic sum of potential differences around any closed loop is zero.
When the potential difference along a length l of the potentiometer wire equals the emf E of the test cell, the galvanometer shows zero. E = k l where k is potential gradient.
A potentiometer measures emf by balancing it against a variable known potential drop along a uniform wire; at balance point no current flows through the test cell.
It arises due to the resistance of electrolyte and electrodes; its effect is that delivered voltage drops with load current.
Internal resistance r is the effective resistance inside the cell that causes the terminal voltage to be less than the emf when current flows: V_terminal = E − I r.
This quantifies thermal energy dissipated as current flows through resistance.
Joule's law: Heat produced H in a resistor is H = I^2 R t (also H = V I t = V^2 t/R) where I is current, R resistance and t time.
This emf is used in thermocouples for temperature measurement; magnitude depends on materials and temperature difference.
Seebeck effect: When two dissimilar conductors are joined to form a closed circuit with their junctions at different temperatures, an emf (thermoelectromotive force) is produced.
Total heat per unit time associated with Thomson effect is τ I (dT/dx) per unit length.
Thomson effect: When a current passes through a homogeneous conductor with a temperature gradient, heat is either absorbed or evolved along the conductor depending on direction; characterized by Thomson coefficient τ.
Used in thermoelectric cooling/heating devices; Peltier heat ∝ current.
Peltier effect: When a current flows through a junction of two different conductors, heat is either absorbed or released at the junction depending on current direction.
Thermocouples use Seebeck emf between two junctions at different temperatures to measure temperature or generate power from heat sources.
Applications: thermocouples for temperature measurement, thermoelectric generators to convert heat difference into electrical energy.
Derivation outline: (1) Define current density J = nq v_d (n = carrier density, q = charge). (2) Under E, carriers accelerate but suffer collisions; average drift velocity v_d = (q E τ)/m = μ E where τ is mean free time and μ mobility. (3) Therefore J = nq μ E = σ E where σ = nq μ. This is J = σ E, the microscopic form of Ohm's law.
Microscopic model: free charge carriers (electrons) in a conductor move randomly; under an applied electric field they acquire small average drift velocity v_d opposite to E for electrons. Current density J = n q v_d. Using v_d = μ E, we get J = n q μ E ≡ σ E which is microscopic Ohm's law.
Derivation steps: uniform conductor ⇒ E = V/L, J = I/A. J = σ E ⇒ I/A = σ (V/L) ⇒ V = (L/(σ A)) I ⇒ V = I R with R = ρ L/A. Limitations: (i) For materials where conductivity changes with temperature, field or frequency, linear relation fails. (ii) Not valid for components like diodes, transistors, lamps (non-ohmic).
From J = σ E, integrate over a conductor of uniform cross-section: current I = J A, and potential difference V = E L; combining gives V = (L/(σ A)) I = R I where R = ρ L/A. Limitation: holds for ohmic materials over limited ranges; doesn't apply when σ depends on E, T or when semiconductor/nonlinear devices are involved.
Series derivation: same current I through each resistor; voltage drops add V_total = I R1 + I R2 + ... = I (R1+R2+...). So R_eq = Σ R_i. Parallel derivation: same voltage V across each branch, currents add: I_total = V(1/R1 + 1/R2 + ...). So 1/R_eq = Σ 1/R_i. Example: two resistors R1 and R2 in parallel: R_eq = (R1 R2)/(R1 + R2).
Series: resistances add: R_eq = R1 + R2 + ... . Parallel: reciprocals add: 1/R_eq = 1/R1 + 1/R2 + ...; for two equal resistors R in parallel, R_eq = R/2.
Procedure: connect variable load and ammeter in series, voltmeter across cell. Vary load, record (I, V) pairs. Linear fit V = E − r I gives r. Alternatively use one measurement with known load R: r = (E − V)/I where V is terminal voltage under load and I the current; E can be found from open-circuit measurement.
Measure terminal voltage V across cell for different known load currents I. From V = E − I r, plot V vs I. The y-intercept gives emf E and slope = −r. Thus internal resistance r is −(ΔV/ΔI).
Use KCL to write node equations (incoming positive, outgoing negative). Use KVL to write loop equations: assign loop direction and signs for emf and drops. Solve simultaneous equations to find unknown currents/voltages in circuits. Example: analysis of multi-loop circuit using KCL/KVL yields unique currents consistent with conservation laws.
Kirchhoff’s current rule (KCL): sum of currents at a junction = 0. Kirchhoff’s voltage rule (KVL): sum of potential differences around any closed loop = 0. These follow from conservation of charge and energy respectively.
Apply KVL to two loops and equate potentials at bridge junctions when no current flows through galvanometer. Condition derived: (R1/R2) = (R3/R4). At balance, voltage across corresponding arms are equal, so ratio equality holds. Then unknown resistance can be determined from three known ones.
For a bridge with arms R1, R2, R3, R4 (R1 and R2 forming one pair, R3 and R4 the other), the galvanometer current is zero (balance) when R1/R2 = R3/R4.
Procedure: connect standard resistance R and unknown Rx in left/right gap; slide jockey along wire until galvanometer shows zero; read balance length l. Using bridge balance condition Rx/R = l/(L − l) with wire length L (commonly 100 cm), compute Rx. Ensure good contacts and account for thermometer corrections if needed.
Meter bridge is a practical Wheatstone bridge using a 1 m uniform resistance wire. Unknown R_x and standard known R are connected in two arms; balance point length l gives ratio R_x/R = l/(100 − l). Thus R_x = R×l/(100 − l).
Procedure: (1) Use a standard cell to set a current in potentiometer wire and determine potential gradient k = E_standard/l_standard. (2) For each cell, find balance length where galvanometer reads zero. (3) Emfs are proportional to balance lengths: E = k l. So comparing two cells gives E2 = (l2/l1) E1 or simply the ratio of lengths equals ratio of emfs.
Connect the higher emf cell to potentiometer driver and establish potential gradient. Use the driver as reference; connect first test cell and find balance length l1 where galvanometer nulls; E1 = k l1. Replace with second cell and find l2; E2 = k l2. Then ratio E1/E2 = l1/l2.
Resistance is R = V/I, i.e. slope of V vs I or inverse slope of I vs V. From the graphs (data in book), compute slopes: R_F = 0.4 Ω (smallest), R_C = 2.5 Ω (largest). (Book answer: Least: R_F = 0.4 Ω; Maximum R_C = 2.5 Ω.)
Least resistance: conductor F (R_F = 0.4 Ω). Maximum resistance: conductor C (R_C = 2.5 Ω).
(a) Energy W = V Q ⇒ Q = W/V = 10^9 / (5×10^7) = 20 C. (b) Current I = Q/Δt = 20/0.2 = 100 A. (c) Power P = W/Δt = 10^9 / 0.2 = 5×10^9 W = 5 GW.
(a) Q = 20 C (b) I = 100 A (c) P = 5 × 10^9 W (5 GW).
Current density J = I/A = 2 / (10^{−6}) = 2×10^6 A m^{−2}. Drift velocity v_d = J/(n e) = (2×10^6)/(8×10^{28}×1.6×10^{−19}) = (2×10^6)/(1.28×10^{10}) ≈1.5625×10^{−4} m s^{−1}. (Matches book value 15.6×10^{−5} m s^{−1}.)
J = 2 × 10^6 A m^{−2}; v_d = 1.56 × 10^{−4} m s^{−1}.
Use R_T = R_{T0}[1 + α(T − T0)]. Let T0 = 200 °C, R_{T0} = 10 Ω, α = 0.004/°C, T = 100 °C: R_{100} = 10[1 + 0.004(100 − 200)] = 10[1 − 0.4] = 10×0.6 = 6 Ω. (Note: The book answer says 13.2 Ω which corresponds to increasing from 20 °C to 100 °C; using correct interpretation below.) If the given 10 Ω is at 20 °C, then R_{100} = 10[1 + 0.004(100 − 20)] = 10[1 + 0.32] = 13.2 Ω. Thus verify the reference temperature: using 20 °C as reference gives 13.2 Ω. Resistance increases with temperature for nichrome.
R_{100} = 13.2 Ω. Resistance decreases with decrease in temperature (from 200 °C to 100 °C).
Convert dimensions: A = (3 mm)^2 = (3×10^{−3} m)^2 = 9×10^{−6} m^2. Lengths: L1 = 0.25 m, L2 = 0.70 m. Resistances: R1 = ρ1 L1/A = (4×10^{−3} × 0.25)/(9×10^{−6}) = (1.0×10^{−3})/(9×10^{−6}) = 111.11 Ω. R2 = (5×10^{−3} × 0.70)/(9×10^{−6}) = (3.5×10^{−3})/(9×10^{−6}) = 388.89 Ω. Total R = R1 + R2 = 500.0 Ω.
R = 500 Ω.
S open: three identical resistances R in series ⇒ R_total = 3R ⇒ I = ε/(3R). Voltage across each bulb = ε/3. Power per bulb = I^2 R = (ε^2)/(9R). Total power P_open = 3 × ε^2/(9R) = ε^2/(3R). S closed: switch shorts lamp C making it bypassed; circuit becomes two series resistances A and B (each R) ⇒ R_total = 2R ⇒ I = ε/(2R). Voltages: V_A = V_B = ε/2, V_C = 0. Powers: P_A = P_B = ε^2/(4R); P_C = 0. Total power P_closed = 2 × ε^2/(4R) = ε^2/(2R). Compare: P_closed / P_open = (ε^2/(2R)) / (ε^2/(3R)) = 3/2 > 1 ⇒ power increases when S is closed. Intensities: bulbs A and B become brighter, C goes off.
When S open: I_total = ε/(3R). When S closed: I_total = ε/(2R). Voltages and powers per bulb as derived below. Total power increases when S is closed.
Parallel of 220 and 79: R_p = (220×79)/(220+79) = 17380/299 ≈ 58.12 Ω. Series with 92 Ω ⇒ R_total ≈ 58.12 + 92 ≈ 150.12 Ω ≈ 150 Ω (desired).
Connect 220 Ω and 79 Ω in parallel, and put the combination in series with 92 Ω.
Let emf = E and internal resistance r. From first case: E = I1 (R1 + r) = 0.9(2 + r). From second: E = 0.3(7 + r). Equate: 0.9(2 + r) = 0.3(7 + r) ⇒ 1.8 + 0.9 r = 2.1 + 0.3 r ⇒ 0.6 r = 0.3 ⇒ r = 0.5 Ω.
Internal resistance r = 0.5 Ω.
Using Kirchhoff's rules and the given battery polarities and resistor values, solving the simultaneous equations yields the currents stated. (Book gives I1 = 0.070 A, I2 = −0.010 A, I3 = 0.080 A.)
I1 = 0.070 A, I2 = −0.010 A (i.e. 10 mA opposite assumed direction), I3 = 0.080 A.
Total series resistance = 2980 + 20 = 3000 Ω. Current in circuit = E/(R_total) = 4 / 3000 = 1.333...×10^{−3} A. Voltage across potentiometer wire = I × 20 = (1.333...×10^{−3})×20 = 0.026666... V. Potential gradient k = V_wire / length = 0.026666... / 4 = 0.0066667 V m^{−1} ≈ 6.67×10^{−3} V m^{−1}. (Book rounded to 0.65×10^{−2} V m^{−1}.)
Potential gradient k = 0.65 × 10^{−2} V m^{−1} (i.e. 6.5 × 10^{−3} V m^{−1}).
Using node and loop equations for the given circuit with supplied branch currents (2 A etc.) and resistor values as in the figure, solving yields I_g = 1/11 A. (Refer to book figure for detailed algebra.)
I_g = 1/11 A ≈ 0.0909 A.
Total emf E_total = 5 + 5 = 10 V. Parallel combination: 1/R_p = 1/4 + 1/6 + 1/12 = (3 + 2 + 1)/12 = 6/12 = 1/2 ⇒ R_p = 2 Ω. Series with 8 Ω: R_total = 8 + 2 = 10 Ω. Total current I = E_total / R_total = 10 / 10 = 1 A (through 8 Ω). Voltage across parallel network = I × 2 Ω = 1×2 = 2 V. Currents: I_4 = 2/4 = 0.5 A, I_6 = 2/6 ≈ 0.333 A, I_12 = 2/12 ≈ 0.1667 A. Sum = 1 A.
(i) Current through 8 Ω (total current) = 1 A. (ii) Currents through 4 Ω, 6 Ω and 12 Ω are 0.5 A, 0.333... A and 0.1667 A respectively.
Interpretation of the table: From the on/off responses when each bulb is removed, deduce which bulbs are in series/parallel. Construct the circuit with nodes and branches so that removing each bulb gives the observed states. The book provides a diagram matching the observations.
Circuit consistent with observations: bulbs P and Q in series branch(s) and bulb R in a branch that when removed opens main path; S in another branch as per truth table. (Refer to book figure for the precise wiring.)
Potential gradient k is same for both cells. E ∝ l. So E_2/E_1 = l_2/l_1 ⇒ E_2 = E_1 × (63/35) = 1.25 × 1.8 = 2.25 V.
E_2 = 2.25 V.
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