What is the answer to the Current Electricity NEET 2026 heater question?
The heater consumes approximately 331 W, option C, when its supply falls from 220 V to 200 V. This Current Electricity NEET 2026 question gives a heater that delivers its rated 400 W at 220 V and asks for its approximate power consumption at 200 V.
The supplied choices are:
The chapter is Current Electricity, and the topic is Electric Power and Heating Effect of Current. The question bank classifies it as medium, with an expected solving time of 60 seconds, not a measured student average.
How do you write the rated and actual power equations?
Keep the heater’s resistance unchanged, not its power. The 400 W rating tells you the power consumed at the specified voltage of 220 V, not at every supply voltage. The official solution treats the heater resistance as unchanged; real temperature-dependent changes in resistance are outside this question’s model.
Define the rated power, rated voltage and applied voltage:
Let the unknown consumed power be:
At the rated voltage, the power relation is:
At the actual applied voltage, it becomes:
Divide the second equation by the first:
The resistance cancels because the same resistance appears in both equations. Choose the ratio method here: calculating resistance first adds work without helping the main solution. Identify what remains fixed before substituting numbers; the power formula alone does not tell you which quantity the problem holds constant.
How does squaring the voltage ratio give 331 W?
Multiply the rated power by the square of the actual-to-rated voltage ratio. This follows directly from dividing the two power equations. Keep the voltage ratio as a fraction until the final division to avoid rounding early.
Rearrange the power ratio:
Substitute the given values:
Simplify first, then square:
Therefore:
Final answer: option C. Check the direction and size of the change before marking it:
The voltage decreases by approximately 9.1%, while power decreases by approximately 17.4%. Power must be below 400 W, and its percentage decrease is larger because voltage is squared under the fixed-resistance model.
Use this repeatable sequence:
- Identify the fixed resistance.
- Write the power ratio.
- Square the voltage ratio.
- Multiply by the rated power.
Why is option B, 400 W, based on the wrong assumption?
Option B treats the rating as fixed consumption rather than a value tied to operating conditions. The incorrect reasoning is: “The label says 400 W, so the heater must consume 400 W even at 200 V.” That confuses a rating at a specified voltage with a conserved or regulated quantity.
The relevant relation is:
With resistance fixed, reducing voltage reduces power. Nothing in the question specifies a controller that maintains constant power.
The other familiar power formula remains valid:
But current also changes according to:
You cannot silently hold current or power constant while reducing voltage. Resistance is the fixed quantity in this model; voltage, current and power change. Using the current formula correctly gives the same voltage-squared dependence, because both voltage and current fall.
What three practice questions test the same heater model?
Calculate resistance from the rating, current at the reduced voltage and heat over a stated time. These are original practice questions based on the heater data, not additional NEET PYQs. Keep the same fixed-resistance assumption used in the Current Electricity NEET 2026 solution.
What resistance follows from the 400 W, 220 V rating?
The heater’s resistance is 121 ohms. Here, calculating resistance is the task itself, so rearrange the rated-power equation rather than using a power ratio.
Use the rated voltage with the rated power. Pairing the reduced voltage with 400 W would mix values from different operating conditions.
What current flows at 200 V if resistance stays unchanged?
The current is approximately 1.65 A. Apply Ohm’s law using the reduced supply voltage and the resistance just obtained.
This current belongs to operation at 200 V. The rating does not force the heater to draw the same current at every voltage.
How much heat is produced at 200 V in 60 seconds?
Approximately 19.8 kJ of electrical energy is converted to heat. For this new exercise, choose an operating interval of 60 s. This is an added condition, not a duration supplied in the original heater question.
Use the unrounded power and round only at the end to avoid unnecessary error. Power measures the rate of energy conversion in watts; heat or energy is measured in joules.
For your next attempt, cover the working and write the fixed quantity beside the given values before selecting a formula.
Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).
Related on NEET JEEnius AI: Work Energy Power Practice Questions NEET: 6 Solved MCQs.
Frequently asked questions
What is the answer to the Current Electricity NEET 2026 heater question?
The heater consumes approximately 331 W, so the answer is option C. With resistance unchanged, its power at 200 V is 400 × (200/220)² = 330.58 W.
Which formula gives heater power when voltage changes?
For unchanged resistance, use P = P₀(V/V₀)², where P₀ is the rated power, V₀ is the rated voltage and V is the applied voltage. This follows from P = V²/R by dividing the actual-power equation by the rated-power equation.
Why does a 400 W heater not consume 400 W at 200 V?
The 400 W rating applies at the specified voltage of 220 V, not at every supply voltage. Under the question's fixed-resistance model, reducing voltage reduces both current and power. No constant-power controller is specified.
How much heat does the heater produce at 200 V in 60 seconds?
Assuming unchanged resistance, the heater converts approximately 19.8 kJ of electrical energy into heat in 60 seconds. Use H = Pt = (40000/121) × 60 J and round only at the end. The 60-second interval is an added practice condition, not part of the original question.