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Electrostatics & Electricity
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Q.1
WBCS Prelims 2004
What is stored in a storage cell?
A. Electric charge
B. Electric potential
C. Lead or some other metal
D. Chemical energy
Explanation
Why Correct: Storage cells (rechargeable batteries) fundamentally store chemical energy that converts to electrical energy during discharge.
Distractor Analysis: Electric charge accumulates on electrodes during charging but represents energy transfer, not storage. Electric potential (voltage) is the energy per charge, not the stored quantity. Lead is an electrode material in lead-acid batteries, not what the cell stores.
Takeaway: Primary cells (non-rechargeable) also store chemical energy but convert it irreversibly.
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Q.2
WBCS Prelims 2023
Combine three resistors 5Ω, 4.5Ω and 3Ω in such a way that the total resistance of this combination is maximum with value
A. 12.5 Ω
B. 13.5 Ω
C. 14.5 Ω
D. 16.5 Ω
Explanation
Core Formula/Logic: For maximum total resistance, connect all resistors in series: R_total = R1 + R2 + R3. For minimum resistance, connect in parallel: 1/R_total = 1/R1 + 1/R2 + 1/R3.
Step-by-Step Solution: 1. To maximize resistance, use series combination.
2. Add all resistances: 5Ω + 4.5Ω + 3Ω = 12.5Ω.
Common Pitfall: Adding incorrectly gives 5+4.5+3=12.5Ω (correct), but misadding as 5+4.5+3=12.5Ω matches option A. Option B (13.5Ω) comes from adding 5+4.5+4, option C (14.5Ω) from 5+4.5+5, option D (16.5Ω) from 5+4.5+7.
Shortcut/Takeaway: For maximum resistance with given resistors, always connect in series and simply sum them. For minimum resistance, use parallel combination and calculate reciprocal sum.
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Q.3
WBCS Prelims 2021
A battery consists of 10 cells, each of emf 1V. If 2 cells are wrongly connected, the emf of the battery becomes
A. 8V
B. 10V
C. 6V
D. 12V
Explanation
Core Formula/Logic: Net emf = (Number of cells in correct polarity - Number of cells in reverse polarity) × emf per cell. Reverse-connected cells subtract their emf from the total.
Step-by-Step Solution: 1. Total cells = 10, each emf = 1V.
2. 2 cells are reverse-connected, so they contribute -2V total.
3. 8 cells are correctly connected, contributing +8V total.
4. Net emf = 8V + (-2V) = 6V.
Common Pitfall: Adding all cells ignoring reverse connection gives 10V (option B). Subtracting reverse cells only once (10-2=8) gives 8V (option A). Adding reverse cells instead of subtracting gives 12V (option D).
Shortcut/Takeaway: For series-connected cells, net emf = (N_correct - N_reverse) × emf_per_cell. Each reverse cell reduces total by twice its individual emf relative to all being correct.
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Q.4
WBCS Prelims 2017
The electric appliances in a house are connected
A. in series
B. in parallel
C. either in series or in parallel
D. both in series and in parallel
Explanation
Why Correct: Household appliances connect in parallel to maintain independent operation at the same voltage (230V in India).
Distractor Analysis: Series connection would cause all appliances to turn off if one fails, and voltage would divide among them. Appliances never use both series and parallel simultaneously in residential wiring. Some industrial or decorative lighting may use series, but not household appliances.
Takeaway: Parallel circuits provide constant voltage across all branches, while series circuits provide constant current through all components.
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Q.5
WBCS Prelims 2014
Basically domestic electric wiring is a
A. parallel connection
B. series connection
C. combination of series and parallel connections
D. None of the above
Explanation
Why Correct: Domestic wiring uses parallel connections to maintain uniform voltage (230V in India) across all appliances, enabling independent operation.
Distractor Analysis: Series connections cause cumulative voltage drops, preventing appliances from receiving full rated voltage. Combination circuits introduce unnecessary complexity and voltage regulation problems in standard home wiring. None of the above is wrong because parallel connection is the fundamental design.
Takeaway: Parallel circuits keep voltage constant across branches while total current equals the sum of individual branch currents.
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Q.6
WBCS Prelims 2012
When a positively charged conductor is earthed, then –
A. Electrons flow from conductor to earth
B. Protoms flow from conductor to earth
C. Electrons flow from earth to conductor
D. Protoms flow from earth to conductor
Explanation
Why Correct: Earth provides an infinite reservoir of electrons that flow to the positively charged conductor to neutralize its charge deficiency.
Distractor Analysis: Electrons flowing from conductor to earth would increase positive charge. Protons cannot flow in conductors as they're bound in atomic nuclei. Protons flowing from earth to conductor would increase positive charge.
Takeaway: For negatively charged conductors, electrons flow from conductor to earth. Current direction convention (positive charge flow) is opposite to actual electron flow direction.
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Q.7
WBCS Prelims 2011
Two copper wires A and B have the same weight and the radius of B is half that of A. The ratio RA/RB of their resistances is
A. 1/4
B. 1/8
C. 1/16
D. 1/2
Explanation
Core Formula/Logic: Resistance R = ρl/A, where ρ is resistivity, l is length, A is cross-sectional area. For same material and weight, volume and thus A*l is constant.
Step-by-Step Solution: 1. Let radius of A = r, radius of B = r/2. 2. Area A_A = πr^2, A_B = π(r/2)^2 = πr^2/4. 3. Same weight implies A_A*l_A = A_B*l_B. 4. So l_A/l_B = A_B/A_A = (πr^2/4)/(πr^2) = 1/4. 5. R_A = ρ*l_A/A_A, R_B = ρ*l_B/A_B. 6. Ratio R_A/R_B = (l_A/A_A)/(l_B/A_B) = (l_A/l_B)*(A_B/A_A) = (1/4)*(1/4) = 1/16.
Common Pitfall: Using only area ratio (1/4) gives 1/4 (option A). Using only length ratio (1/4) gives 1/4 (option A). Multiplying incorrectly as 1/2*1/4 = 1/8 gives option B.
Shortcut/Takeaway: For same material and weight, resistance is proportional to 1/(radius^4). Halving the radius increases resistance 16 times, so the ratio of the thicker wire's resistance to the thinner is 1/16.
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Q.8
WBCS Prelims 2010
In a house-hold wiring the appliances are connected in series.
A. True
B. False
C. In parallel
D. In mixed configuration
Explanation
Why Correct: Household appliances connect in parallel to maintain independent operation and constant voltage across each device.
Distractor Analysis: Series wiring would cause all appliances to fail if one device breaks and would force current through each device sequentially. Parallel wiring allows each appliance to operate independently at full voltage. Mixed configurations exist in complex systems but not in basic household wiring.
Takeaway: Parallel circuits dominate domestic wiring for reliability, while series circuits appear in simple strings like decorative lights.
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Q.9
WBCS Prelims 2010
Electrical energy is converted into mechanical energy by a
A. Thermostat
B. Motor
C. Dynamo
D. Rectifier
Explanation
Why Correct: An electric motor uses electromagnetic forces to convert electrical energy into rotational mechanical energy, powering devices from fans to industrial machinery.
Distractor Analysis: A thermostat regulates temperature by switching circuits on/off. A dynamo (generator) converts mechanical energy into electrical energy, the reverse process. A rectifier converts alternating current to direct current electricity.
Takeaway: Remember the energy conversion pairs: motor (electrical→mechanical), generator/dynamo (mechanical→electrical), microphone (sound→electrical), speaker (electrical→sound).
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Q.10
WBCS Prelims 2008
The resistance of which of the following decreases with the rise of temperature ?
A. Copper
B. Iron
C. Silicon
D. Mercury
Explanation
Why Correct: Silicon is a semiconductor whose resistance decreases with temperature rise because increased thermal energy excites more electrons into the conduction band, increasing conductivity.
Distractor Analysis: Copper, iron, and mercury are metals whose resistance increases with temperature due to enhanced lattice vibrations that scatter conduction electrons more effectively.
Takeaway: Germanium, carbon, and other semiconductors also exhibit negative temperature coefficients of resistance, while metals like aluminum, silver, and gold show positive coefficients.
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Q.11
WBCS Prelims 2008
The resistance of a 200 V-100 W bulb is
A. 400 Ω
B. 400 Ω only when it is connected to 200 volt mains
C. 400 Ω when it is not glowing
D. 2 Ω
Explanation
Core Formula/Logic: Electrical power P = V2/R, where P is power in watts, V is voltage in volts, and R is resistance in ohms.
Step-by-Step Solution: 1. Given: P = 100 W, V = 200 V.
2. Rearrange formula: R = V2/P.
3. Calculate: R = (200*200)/100 = 40000/100 = 400 Ω.
4. This resistance value is the nominal resistance at rated voltage, regardless of connection status.
Common Pitfall: Confusing resistance with power rating leads to R = P/V = 100/200 = 0.5 Ω, which is not among options. Misunderstanding that resistance changes with temperature might lead to selecting option B or C, but the question asks for the resistance of the bulb, which is its rated resistance.
Shortcut/Takeaway: For any bulb rated V volts and P watts, resistance R = V2/P. Memorize: 200V-100W bulb always has R = 400 Ω.
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Q.12
WBCS Prelims 2007
The motion of which particle through a metallic wire is called electric current?
A. Electron
B. Positron
C. Neutron
D. Photon
Explanation
Why Correct: Electric current in metallic conductors consists of electron flow—negatively charged particles moving from lower to higher potential.
Distractor Analysis: Positrons are antimatter particles with positive charge, not present in normal metallic conduction. Neutrons are neutral particles within atomic nuclei, not involved in charge transport. Photons are massless particles of electromagnetic radiation, not charge carriers in wires.
Takeaway: Conventional current direction (positive to negative) opposes actual electron flow direction (negative to positive) in metallic conductors.
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Q.13
WBCS Prelims 2004
If the length and cross-section of a wire are both doubled the resistance will
A. increased 8 times
B. decreases 4 times
C. increase twice
D. remain unchanged
Explanation
Core Formula/Logic: Resistance R = ρL/A, where ρ is resistivity, L is length, A is cross-sectional area.
Step-by-Step Solution: 1. Initial resistance: R1 = ρL/A. 2. New length L' = 2L. 3. New area A' = 2A (since cross-section doubled). 4. New resistance R2 = ρ(2L)/(2A) = ρL/A = R1. 5. Therefore resistance remains unchanged.
Common Pitfall: Multiplying only length gives R2 = ρ(2L)/A = 2R1, which produces option C. Multiplying only area gives R2 = ρL/(2A) = R1/2, which produces option B. Multiplying both incorrectly as 2L × 2A = 4 gives R2 = ρ(4L)/A = 4R1, which produces option A.
Shortcut/Takeaway: When both length and area change by factor k, resistance changes by factor k/k = 1. So doubling both cancels out exactly.
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Q.14
WBCS Prelims 2004
The specific resistance of a conductor depends on its
A. length
B. width
C. temperature
D. shape of the cross-section
Explanation
Why Correct: Specific resistance (resistivity) is an intrinsic property of the material, defined as ρ = R × A / L, where R is resistance, A is cross-sectional area, and L is length. It depends on temperature because atomic vibrations increase with temperature, scattering electrons more and increasing resistivity.
Distractor Analysis: Length and width affect the conductor's resistance but not its specific resistance. Shape of the cross-section affects area but not the material's intrinsic resistivity.
Takeaway: For most metals, resistivity increases linearly with temperature: ρ = ρ0[1 + α(T - T0)], where α is the temperature coefficient of resistivity.
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Q.15
WBCS Prelims 2004
The only vector quantity among the following is
A. electric charge
B. electric potential
C. electric field intensity
D. electric resistance
Explanation
Why Correct: Electric field intensity (E = F/q) is a vector quantity with both magnitude and direction, measured in volts/meter.
Distractor Analysis: Electric charge is a scalar quantity measured in coulombs. Electric potential is a scalar quantity measured in volts. Electric resistance is a scalar quantity measured in ohms.
Takeaway: In electromagnetism, other vector quantities include magnetic field intensity, current density, and Poynting vector, while scalar quantities include electric flux, capacitance, and inductance.
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Q.16
WBCS Prelims 2004
The current on a 100W, 220V electric bulb is
A. 2.2 amp.
B. 1.1 amp.
C. 5/11 amp.
D. 22000 amp.
Explanation
Core Formula/Logic: Power formula: I = P ÷ V, where I is current in amperes, P is power in watts, and V is voltage in volts.
Step-by-Step Solution: 1. Given P = 100 W, V = 220 V.
2. Apply formula: I = 100 ÷ 220.
3. Simplify: I = 10 ÷ 22 = 5 ÷ 11 A.
Common Pitfall: Dividing 220 by 100 gives 2.2 A (option A). Forgetting to divide gives 22000 A (option D). Incorrect simplification yields 1.1 A (option B).
Shortcut/Takeaway: For quick mental calculation, note that 100/220 = 10/22 = 5/11 ≈ 0.4545 A. Memorize that I = P/V always gives current in amperes when power is in watts and voltage in volts.
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Q.17
WBCS Prelims 2001
The emf of a cell does not depend on
A. the size of the cell
B. the material of cathode
C. the material of anode
D. electrolyte used
Explanation
Why Correct: Electromotive force (emf) depends on the electrode materials and electrolyte composition, which determine the standard electrode potentials and concentration differences, but not on the physical dimensions of the cell.
Distractor Analysis: The material of cathode and anode directly determines the standard reduction potentials that contribute to the cell potential. The electrolyte used affects the concentration and activity of ions, influencing the Nernst equation calculation of emf. The size of the cell only affects current capacity and internal resistance, not the thermodynamic potential difference.
Takeaway: For a galvanic cell, Ecell = Ecathode - Eanode under standard conditions, and E = E0 - (RT/nF)lnQ under non-standard conditions, where Q depends on electrolyte concentrations.
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