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Test: Ohm's Law

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5 Questions MCQ Test | Test: Ohm's Law

Test: Ohm's Law for Class 10 2023 is part of Class 10 preparation. The Test: Ohm's Law questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Ohm's Law MCQs are made for Class 10 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Ohm's Law below.
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Test: Ohm's Law - Question 1

The water heating rod draws 10 A current when connected to certain power source. The resistance of rod is 12 Ω. The source voltage is:

Detailed Solution for Test: Ohm's Law - Question 1

The correct answer is B as Ohm's law:
voltage=current x resistance=12x10=120 V

Test: Ohm's Law - Question 2

Which method can be used for absolute measurement of resistances ?

Detailed Solution for Test: Ohm's Law - Question 2

A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. The primary benefit of the circuit is its ability to provide extremely accurate measurements.

Test: Ohm's Law - Question 3

Electric current always flows

Detailed Solution for Test: Ohm's Law - Question 3

Electrons move from low potential to high potential. The direction of electric current is opposite to the direction of flow of electrons.

Hence Electric current always flows from higher to lower potential.

Test: Ohm's Law - Question 4

What is the kilowatt-hour consumption of a 40 W lamp if it remains on for 1750 h ?

Detailed Solution for Test: Ohm's Law - Question 4

(1,750 x 40)/1000 = 70 kilowatt-hour.

Test: Ohm's Law - Question 5

The potential difference of a circuit is constant. If the resistance of a circuit is doubled, then its current will become;

Detailed Solution for Test: Ohm's Law - Question 5

According to ohm's law

Current (I) = Potential difference (V) / Resistance(R)

If the potential difference is maintained constant and the resistance is changed,

Cureent (I) is inversely proportinal to the reisistance (R)

V = IR                     --(1) 

Now, here V is constant.

Let , new resistance R' = 2R and assume that current flow be I'

So, from (1),

IR = I'. R',

=> IR = I'. 2R,

Therefore, I' = I / 2

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