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Electrical energy and power, electrical resistivity and conductivity MCQs for JEE

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Electrical energy and power, electrical resistivity and conductivity JEE MCQs on Goodmarks include 30+ multiple-choice questions with correct answers and step-by-step solutions. Attempt free samples below or unlock the full bank with Pro.

Master Electrical energy and power, electrical resistivity and conductivity through exam-style multiple-choice questions. This page features 30+ JEE MCQs covering Electrical energy and power, electrical resistivity and conductivity, each with verified answers and clear explanations.

Syllabus context

Part of Current Electricity in JEE Main Physics.

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Q1PhysicsUnit 12: Current Electricity
In the circuit shown, the heat produced in 5Ω5 \Omega resistor is 10 calorie/sec. The heat produced/sec in 4Ω4 \Omega resistor will be :
Q2PhysicsUnit 12: Current Electricity
Q Type your question- and of fixed emf, the resistor R1\boldsymbol{R}_{1} has fixed resistance and the resistance of resistor R2\boldsymbol{R}_{2} can be varied (but R2\boldsymbol{R}_{2} is always nonzero). Then the electric power delivered to the resistor of resistance R1R_{1} is independent of the value of resistance R2\boldsymbol{R}_{2} Reason f potential difference across a fixed resistance is unchanged, the power delivered to the resistor remains constant.
Q3PhysicsUnit 12: Current Electricity
The most important safety method used for protecting home appliances from short circuiting or overloading is
Q4PhysicsUnit 12: Current Electricity
The electrical energy in kilowatt hours consumed in operating ten 50 W50 ~ W bulbs for 10 hrs/day in a month of 30 days is
Q5PhysicsUnit 12: Current Electricity
The Joules heating effect for an electric heater in A.CA . C and in D.C. circuit is
Q6PhysicsUnit 12: Current Electricity
The heat developed in a system is proportional to the current through it. Then
Q7PhysicsUnit 12: Current Electricity
A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be:
Q8PhysicsUnit 12: Current Electricity
P=V2RP=\frac{V^{2}}{R} is applied when

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JEE Main tests Electrical energy and power, electrical resistivity and conductivity through conceptual and numerical MCQs. Goodmarks mirrors this format with four-option questions and detailed solutions.

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