CBSE Class 10 Science – Chapter Wise Study Materials

Case 1 — Ohm’s Law in a wire

Topic: Ohm's law and linearity

Scenario: A student connects a wire across a variable DC source and measures current for different voltages. She finds that when V = 2.0 V, I = 0.50 A; when V = 6.0 V, I = 1.50 A.

Questions:

Is the wire obeying Ohm’s law? Justify using the data.
Calculate the resistance of the wire.

Show Answer

(1) Ohm’s law states V ∝ I (constant R). Check ratio V/I: for first point 2.0/0.50 = 4.0 Ω; second point 6.0/1.50 = 4.0 Ω. Since V/I is constant, the wire obeys Ohm’s law (it is ohmic) under these conditions.
(2) Resistance R = V/I = 4.0 Ω.

Case 2 — Two resistors in series

Topic: Series combination of resistors

Scenario: Two resistors, R₁ = 6 Ω and R₂ = 3 Ω, are connected in series to a 9 V battery.

Questions:

Find the equivalent resistance and current through the circuit.
Calculate the potential difference across each resistor.

Show Answer

Equivalent resistance R_eq = R1 + R2 = 6 + 3 = 9 Ω.
Current I = V / R_eq = 9 V / 9 Ω = 1.0 A (same through both resistors in series).
Potential difference: V1 = I·R1 = 1×6 = 6 V; V2 = I·R2 = 1×3 = 3 V. (Check: V1+V2 = 9 V).

Case 3 — Parallel branches

Topic: Parallel combination and current division

Scenario: Two resistors, 4 Ω and 12 Ω, are connected in parallel across a 12 V supply.

Questions:

Find the equivalent resistance.
Calculate current through each resistor and total current drawn from the supply.

Show Answer

Equivalent resistance: 1/R_eq = 1/4 + 1/12 = (3+1)/12 = 4/12 → R_eq = 12/4 = 3 Ω.
Voltage across each branch is same = 12 V. Currents: I1 = 12/4 = 3 A; I2 = 12/12 = 1 A. Total current I_total = I1 + I2 = 4 A. (Check: V / R_eq = 12 / 3 = 4 A).

Case 4 — Changing a single resistor

Topic: Effect on circuit when one resistor changes

Scenario: In a series circuit with three equal resistors (each R), total current is I. One resistor is replaced by another of resistance 2R. Battery voltage remains same.

Questions:

How does total current change (increase/decrease)? Explain qualitatively.
If initial I = 2 A and each original R = 5 Ω, find new current.

Show Answer

(1) Replacing one resistor by a larger value increases total resistance → current decreases.
(2) Initial R_total = 3R = 3×5 = 15 Ω → I_initial = V / 15 = 2 A → V = 2×15 = 30 V. New R_total = R + R + 2R = 4R = 20 Ω. New current I_new = V / 20 = 30 / 20 = 1.5 A.

Case 5 — Ohmmeter and internal resistance

Topic: Measuring resistance and internal resistance concept

Scenario: A student measures the terminal voltage of a battery while a small resistor is connected and notes a drop from the open-circuit emf. The battery's emf is 12 V; with a load of 4 Ω the terminal voltage is 11.6 V.

Questions:

Find the current through the circuit when 4 Ω is connected.
Calculate the internal resistance of the battery.

Show Answer

Terminal voltage V_term = 11.6 V. Current I = V_term / R_load = 11.6 / 4 = 2.9 A.
Emf E = V_term + I·r → r = (E − V_term) / I = (12 − 11.6) / 2.9 = 0.4 / 2.9 ≈ 0.1379 Ω ≈ 0.14 Ω.

Case 6 — Heating effect and time

Topic: Heating effect of current (Joule heating)

Scenario: An electric iron of resistance 20 Ω is connected to a 220 V supply and used for 30 minutes.

Questions:

Find the current through the iron.
Calculate energy converted to heat (in kWh) in 30 minutes.

Show Answer

Current I = V / R = 220 / 20 = 11 A.
Power P = V·I = 220 × 11 = 2420 W = 2.42 kW. Time t = 30 min = 0.5 h. Energy = P×t = 2.42 × 0.5 = 1.21 kWh.

Case 7 — Two bulbs with different power ratings

Topic: Power rating and brightness

Scenario: Two bulbs, A (60 W, 240 V) and B (100 W, 240 V), are connected—first in series and then individually across a 240 V mains supply.

Questions:

Which bulb is brighter when each is connected individually to 240 V? Why?
If connected in series to 240 V, which bulb glows brighter? Explain.

Show Answer

(1) Individually: Bulb with higher wattage (100 W) dissipates more power at the same voltage → B is brighter.
(2) In series: same current flows through both. Using P = I²R, brightness depends on resistance. From rating R = V²/P: R_A = 240²/60 = 960 Ω; R_B = 240²/100 = 576 Ω. Higher resistance (A) gets larger share of voltage and dissipates more power at same current → bulb A (60 W rated) will be brighter in series.

Case 8 — Faulty appliance and fuse

Topic: Safety devices — fuse and current rating

Scenario: A fan rated at 220 V and 100 W is connected to mains. Its starting current occasionally spikes to 2 A. Choose a fuse for the fan.

Questions:

Find the normal operating current.
Suggest a suitable fuse rating and explain reasoning.

Show Answer

Normal current I = P/V = 100 / 220 ≈ 0.455 A ≈ 0.46 A.
Starting spike = 2 A. Fuses should protect against prolonged overload but allow harmless short spikes. Use a fuse slightly above operating current, commonly 1 A fuse would blow at small overloads; because starting spike is 2 A, choose a slow-blow/thermal fuse with rating around 1 A–1.5 A or a fast fuse of 2 A depending on safety norms. Practical recommendation: use a 1 A slow-blow fuse to tolerate brief surges but protect against sustained high current. (Follow appliance standards and manufacturer guidance.)

Case 9 — Wire length and resistance

Topic: Dependence of resistance on dimensions/material

Scenario: Two wires of identical material and thickness — wire X is 2 m and wire Y is 4 m. Both have the same cross-sectional area.

Questions:

Compare their resistances.
If X has resistance 1 Ω, what is resistance of Y? Explain.

Show Answer

Resistance R ∝ length. Since Y is twice as long, R_Y = 2 × R_X. If R_X = 1 Ω, then R_Y = 2 Ω.

Case 10 — Combining series & parallel

Topic: Mixed circuits

Scenario: A 12 V battery is connected to R1 = 6 Ω in series with a parallel combination of R2 = 3 Ω and R3 = 6 Ω. (So R1 in series with (R2 || R3)).

Questions:

Find equivalent resistance of the whole circuit.
Find current through R1 and through R2.

Show Answer

First compute R2||R3: 1/Rp = 1/3 + 1/6 = (2+1)/6 = 3/6 → Rp = 2 Ω. Total R_eq = R1 + Rp = 6 + 2 = 8 Ω. Current from battery I = 12 / 8 = 1.5 A (this is current through R1). Voltage across parallel block V_p = I × Rp = 1.5 × 2 = 3 V. Current through R2: I2 = V_p / 3 = 3 / 3 = 1 A. (Current through R3 = 3/6 = 0.5 A; 1 + 0.5 = 1.5 A flows through R1 — consistency check).

Case 11 — Electric power at different voltages

Topic: Power calculations and device selection

Scenario: A heater is rated 1500 W at 230 V. A student wants to know the current it draws and what would happen if connected to 115 V supply (half voltage).

Questions:

Calculate normal operating current at 230 V.
If the heater is supplied 115 V and its resistance remains nearly constant, estimate the power at 115 V.

Show Answer

I = P / V = 1500 / 230 ≈ 6.52 A.
Resistance R = V² / P = 230² / 1500 ≈ 35.27 Ω. At 115 V, power P' = V'² / R = 115² / 35.27 ≈ (13225 / 35.27) ≈ 375 W (approx). So at half voltage power reduces to about one-quarter (since P ∝ V² for fixed R).

Case 12 — Ammeter and voltmeter placement

Topic: Correct connection of meters and effect on readings

Scenario: A learner accidentally connects an ammeter in parallel with a resistor and a voltmeter in series. Describe what happens.

Questions:

Predict the readings and dangers for the ammeter.
What reading would the voltmeter show when placed in series?

Show Answer

(1) Ammeter has very low internal resistance and is meant to be in series. If placed in parallel across a resistor, almost all current will pass through the ammeter causing a dangerously large current — it may be damaged or fuse. Reading would be erroneously high (close to total source current) and unsafe.
(2) Voltmeter has high resistance and is meant to be in parallel. In series, it will limit current drastically, so circuit current will be very small; the voltmeter will show only the small drop across itself or possibly the full supply depending on arrangement — the circuit will not function normally. Always connect ammeter in series and voltmeter in parallel.

Case 13 — Source with internal resistance powering a load

Topic: Maximum power transfer idea (qualitative)

Scenario: A source of emf E and internal resistance r is connected to a variable load R. Student varies R and notices power in the load peaks at R = r.

Questions:

Explain why power in the load is maximum when R = r (qualitative).
If E = 12 V and r = 2 Ω, what load gives maximum power and what is that maximum power?

Show Answer

(1) Power delivered to load P = I²R where I = E/(R+r). So P = [E² R] / (R+r)². This expression is maximized when R = r (calculus or inspection). Physically, when R = r, voltage division and current balance produce largest power transfer to R.
(2) With E=12 V, r=2 Ω → optimal R = 2 Ω. Maximum power P_max = E²/(4(r))? Common formula: P_max = E²/(4(R+r)?) — simpler: P_max = (E²) / (4r) when R = r. So P_max = 12² / (4×2) = 144 / 8 = 18 W.

Case 14 — Short circuit scenario

Topic: Short circuit and effects

Scenario: A wire accidentally connects the two terminals of a battery directly (close to zero resistance). Explain consequences.

Questions:

What happens to current and why is it dangerous?
How does internal resistance r affect the short-circuit current?

Show Answer

(1) With very low external resistance, current I ≈ E / r (large). Huge current can cause heating, sparks, fire, battery damage, or explosion — dangerous.
(2) Short-circuit current is mainly limited by internal resistance r because external R ≈ 0. So I_short ≈ E / r. Smaller r → larger dangerous current.

Case 15 — Meter reading in mixed circuit

Topic: Reading current/voltage in series-parallel networks

Scenario: In a circuit, ammeter A is placed in one branch of a parallel network. Branch 1 has two series resistors 2 Ω and 4 Ω; Branch 2 has a single 3 Ω resistor. Supply is 9 V. Ammeter A measures current through branch 1 (through both series resistors).

Questions:

Find the total current from the supply.
What would the ammeter reading be?

Show Answer

Branch 1 resistance R1 = 2 + 4 = 6 Ω → I1 = 9 / 6 = 1.5 A (this is ammeter reading). Branch 2 resistance R2 = 3 Ω → I2 = 9 / 3 = 3 A. Total supply current = I1 + I2 = 4.5 A.

Case 16 — Effect of temperature on resistance

Topic: Temperature dependence of resistance

Scenario: A metal wire’s resistance increases with temperature. A filament lamp heats up when current flows.

Questions:

How does the current change if the applied voltage is kept constant and the filament heats up?
Explain why filament lamps draw less current than expected from cold resistance.

Show Answer

(1) If R increases (due to heating) while V is constant, current I = V/R decreases.
(2) Filament at room temperature has much lower resistance (cold resistance). When switched on, filament heats up rapidly and resistance rises, so steady-state current is lower than the initial surge. That's why cold filament may cause a brief higher inrush current.

Case 17 — Energy cost calculation

Topic: Electrical energy, billing (kWh)

Scenario: A household uses a 60 W LED lamp for 5 hours each day and a 1.5 kW heater for 2 hours daily. Electricity tariff = ₹8.50 per kWh. Estimate daily cost for these two appliances.

Questions:

Compute daily energy consumption (in kWh).
Compute daily cost.

Show Answer

LED energy = 60 W × 5 h = 300 Wh = 0.3 kWh. Heater energy = 1500 W × 2 h = 3000 Wh = 3.0 kWh. Total = 0.3 + 3.0 = 3.3 kWh/day.
Daily cost = 3.3 × ₹8.50 = ₹28.05 per day (approx).

Case 18 — Identical bulbs on different circuits

Topic: Parallel circuits in household wiring

Scenario: Two identical bulbs are connected to the same mains supply — one is switched on and the other off. If one bulb fuses (burns out), what happens to the other bulb?

Questions:

Does the functioning bulb get affected? Explain with circuit-type reason.
If bulbs were instead wired in series, what would happen when one fuses?

Show Answer

(1) In household wiring bulbs are in parallel. If one fuses (open circuit), the other still receives full supply voltage and remains unaffected.
(2) In series, if one fuses (open), the series circuit breaks so the other bulb will also go off.

Case 19 — Determining unknown resistor using circuit measurements

Topic: Use of measured voltage & current to find resistance

Scenario: A student measures voltage across an unknown resistor as 4.5 V and the current through it as 0.15 A.

Questions:

Calculate the resistance.
If the resistor dissipates heat, what is the power converted per second?

Show Answer

R = V / I = 4.5 / 0.15 = 30 Ω.
Power P = V × I = 4.5 × 0.15 = 0.675 W.

Case 20 — Designing a simple resistor divider

Topic: Voltage division and practical application

Scenario: A student needs 6 V from a 12 V source to power a small sensor. She uses two resistors R1 and R2 in series across 12 V and takes the output across R2 (voltage divider).

Questions:

What ratio R1:R2 will give 6 V across R2?
If R1 + R2 should be about 10 kΩ to limit current, choose practical resistor values.

Show Answer

(1) For equal split (6 V from 12 V) we require R1 = R2 (since V_R2 = 12 × R2/(R1+R2)). So ratio R1:R2 = 1:1.
(2) If R_total = 10 kΩ and R1 = R2, choose R1 = R2 = 5 kΩ (standard value: 4.7 kΩ is common — using 4.7 kΩ each gives V_R2 ≈ 12×4.7/(4.7+4.7) = 6 V approximately).

Electricity – Case-based Questions with Answers

Physics — Chapter 11: Electricity

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