Test: Ohm's Law


5 Questions MCQ Test Science Class 10 | Test: Ohm's Law


Description
This mock test of Test: Ohm's Law for Class 10 helps you for every Class 10 entrance exam. This contains 5 Multiple Choice Questions for Class 10 Test: Ohm's Law (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Ohm's Law quiz give you a good mix of easy questions and tough questions. Class 10 students definitely take this Test: Ohm's Law exercise for a better result in the exam. You can find other Test: Ohm's Law extra questions, long questions & short questions for Class 10 on EduRev as well by searching above.
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:

Solution:

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

QUESTION: 2

Which method can be used for absolute measurement of resistances ?

Solution:

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.

QUESTION: 3

Electric current always flows

Solution:

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.

QUESTION: 4

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

Solution:

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

QUESTION: 5

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

Solution:

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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