Ohm’s Law – Long Answer Questions (CBSE Class 10 Science – Physics)

Medium Level (Application & Explanation)

Q1. State Ohm’s law. Write its mathematical form, the conditions for validity, and illustrate its use with a simple real-life example.

Answer: Ohm’s law states that the current (I) through a conductor is directly proportional to the potential difference (V) across its ends, provided temperature and physical conditions remain constant. Mathematically, V = IR, where R is the resistance of the conductor. This relation assumes the conductor does not heat up significantly and that its material and dimensions do not change during measurement. A simple example is a torch bulb at the moment you switch it on: if you double the number of cells, the voltage doubles, so the current also roughly doubles (initially), making the bulb brighter. In copper wire, if you apply 3 V and get 0.3 A, then the resistance is R = V/I = 3/0.3 = 10 Ω. If you increase voltage to 6 V, current becomes 0.6 A, confirming proportionality and constant resistance under the same conditions.

Q2. Using V = IR, explain how to determine current and resistance in different cases. Solve two examples and interpret the results.

Answer: To use Ohm’s law, rearrange the formula as needed:

To find current: I = V/R
To find resistance: R = V/I Example 1: A resistor of R = 20 Ω is connected to a 10 V battery. Then I = 10/20 = 0.5 A. This tells us the circuit will carry a moderate current and the resistor will limit the flow to protect components.
Example 2: A bulb draws 0.3 A when connected to 6 V. Then R = 6/0.3 = 20 Ω. This means, under those conditions, the bulb’s effective resistance is 20 Ω. These calculations help in choosing safe components, predicting current draw, and ensuring no overheating. If, after increasing voltage, current rises in the same ratio, resistance stays constant and the component likely obeys Ohm’s law within that range.

Q3. Describe a classroom activity to verify Ohm’s law. Include the circuit, steps, observations, graph, and conclusion.

Answer: Set up a circuit with a battery, a resistor (wire/coil), an ammeter in series, a voltmeter across the resistor, a switch, and optionally a rheostat. Steps:

Close the switch and record the first pair of readings (V1, I1).
Change the voltage (by battery taps or rheostat) and note (V2, I2), then (V3, I3), ensuring the resistor doesn’t overheat.
Tabulate V and I, compute V/I each time.
Observations: For an ohmic conductor, V/I remains nearly constant; plotting V on y-axis versus I on x-axis gives a straight line through the origin.
Conclusion: The linear V–I graph and constant slope (V/I) show constant resistance and verify Ohm’s law at constant temperature. Precautions: Use short intervals to avoid heating, ensure tight connections, and zero the meters before starting. This activity builds strong, hands-on understanding.

Q4. Explain the V–I graph for an ohmic conductor and show how to compare resistances of two materials using their slopes.

Answer: For an ohmic conductor, the V–I graph (V on y-axis, I on x-axis) is a straight line through the origin. The slope = V/I = R, so the slope equals resistance. To compare two materials, draw both lines on the same axes under identical conditions. The material with the greater slope has higher resistance because more voltage is needed for the same current. For example, nichrome typically has a higher resistance than copper, so its line is steeper, while copper’s line is less steep (lower slope), indicating lower resistance. If a graph is curved, the resistance is not constant (non-ohmic behavior), often due to heating or material properties. This method makes it easy to visualize and rank resistances without complex calculations.

Q5. List practical applications of Ohm’s law in everyday circuits and explain one with a short calculation.

Answer: Ohm’s law is used for:

Designing circuits (e.g., choosing resistors for LEDs so they do not burn out).
Home wiring safety (selecting wire gauges to keep current within safe limits).
Measuring resistance using voltmeters and ammeters.
Selecting fuses with appropriate current ratings to protect appliances.
Adjusting brightness by adding series resistance to lamps.
Example (LED protection): Suppose a 9 V supply powers an LED that needs 2 V at 20 mA. The resistor must drop 9 − 2 = 7 V at 0.02 A. Using R = V/I, we get R = 7/0.02 = 350 Ω. We choose the nearest standard value, say 360 Ω. This simple use of V = IR prevents excess current, protecting the LED and ensuring reliable operation.

Q6. What are the limitations of Ohm’s law? Explain the conditions under which it is valid and give examples of non-ohmic devices.

Answer: Ohm’s law holds only when temperature and physical conditions remain constant and for materials that show a linear V–I relationship. It is limited by:

Heating effects: As a filament bulb heats up, its resistance increases, so the V–I graph becomes curved.
Material type: Semiconductors and diodes are non-ohmic; their current depends on voltage in a nonlinear way, often with a threshold.
Electrolytes and gases: Conduction involves ions and complex interactions, not a simple V–I proportionality.
Mechanical changes: Stretching or compressing a conductor changes its resistance.
Validity: For metals like copper under constant temperature, short measurement intervals to avoid heating, and stable contacts, V ∝ I works well. Recognizing where Ohm’s law applies prevents calculation errors and improves circuit design.

High Complexity (Analytical & Scenario-Based)

Q7. A student records the following for a wire: (V, I) = (2 V, 0.1 A), (4 V, 0.2 A), (6 V, 0.3 A). Analyze if the wire is ohmic, calculate resistance, and discuss possible experimental errors.

Answer: The data show that when voltage doubles, current also doubles. The ratio V/I is 20 Ω in each case, indicating constant resistance and a straight-line V–I graph through the origin. Thus, the wire is ohmic under these conditions, with R = 20 Ω. Possible experimental errors include:

Heating of the wire, which could raise resistance if measurements are too slow.
Contact resistance at loose connections, adding to measured values.
Meter least count and zero error, making small currents/voltages inaccurate.
Parallax while reading analog meters.
Battery internal resistance changing with load, slightly affecting V.
To minimize errors: keep readings quick, tighten connections, zero meters, and, if possible, use a rheostat for smooth adjustments. The consistent ratio here supports Ohm’s law robustly.

Q8. Two wires, A and B, are tested under identical conditions. Their V–I graphs are straight lines through the origin. Wire A’s graph has a higher slope than wire B’s. Compare their resistances, currents at 6 V, and suggest a suitable use for each.

Answer: Since the slope (V/I) equals resistance, a higher slope means higher resistance. Therefore, R_A > R_B. At 6 V, current is I = V/R, so I_A < I_B. In practice, wire A (higher resistance) will allow less current for the same voltage and can be useful where current limiting or heating control is needed (e.g., in a resistive element or for protection in simple setups). Wire B (lower resistance) is better for conducting power efficiently with less voltage drop, such as interconnections or low-loss leads. This comparison shows how reading the slope of straight V–I lines helps choose the right wire for either efficient conduction (low R) or controlled current (high R), all derived directly from V = IR.

Q9. A 12 V lamp behaves approximately ohmically at room temperature with an effective resistance of 6 Ω. A 3 Ω resistor is added in series. Calculate the new current and explain the change in brightness using Ohm’s law reasoning.

Answer: Initially, the lamp’s current is I_initial = V/R = 12/6 = 2 A. After adding a 3 Ω series resistor, the total resistance becomes R_total = 6 + 3 = 9 Ω. The new current is I_new = 12/9 ≈ 1.33 A. Since the same current flows through series components, the lamp now gets only 1.33 A instead of 2 A. By Ohm’s law, increased total resistance reduces current, so the lamp dims. Also, the voltage across the lamp is now V_lamp = I_new × 6 ≈ 1.33 × 6 ≈ 8 V, lower than before, further reducing its output. While exact brightness also depends on filament heating (which can change resistance), the core reason for dimming here is the reduced current due to added series resistance.

Q10. Compare V–I characteristics of an ohmic resistor, a filament bulb, and a diode. Explain the shape of each curve and relate it to Ohm’s law.

Answer:

Ohmic resistor: Shows a straight line through the origin in a V–I graph. The slope (V/I) is constant and equals resistance (R). This obeys Ohm’s law because current is directly proportional to voltage at constant temperature.
Filament bulb: The V–I curve is nonlinear and curves because as current increases, the filament heats up, raising its resistance. For equal voltage steps, current does not increase proportionally. Hence, it deviates from Ohm’s law.
Diode: The V–I graph shows very little current until a threshold/forward voltage is reached, and then current rises rapidly; in reverse bias, current is nearly zero (until breakdown). This is highly nonlinear, so Ohm’s law does not apply in ...