D. P. P. Electricity

D.P.P. 1                                                2-July-2020

◾️Q 1. What is meant by electric current? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current? Give justification for your answer.
A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flows through any section of the conductor in 1 second. (Charge on electron 1.6 X 10-19 coulomb).

◽️ANSWER:- ELECTRIC CURRENT :Electric current is the amount of charge flowing through a conductor per unit time. OR , 

We can say that ELECTRIC CURRENT is the rate of flow of charge.

➡️Its SI unit is AMPERE.

WHAT IS 1 AMPERE -> The current flowing through conductor is said to be 1 Ampere if 1 coloumb of charge passes through it in 1 sec.

Current = Charge / Time

1 Ampere = 1 coloumb / 1 second

➡️ In a conductor, if electrons are flowing from B to A, then the direction of conventional current is from A to B.

➡️Justification -> Because, the direction of conventional current is considered opposite to the flow of electrons. Electric current comprises of the flow of positive charge. So, its direction will be opposite to the direction of negative charges that are electrons.

1 Ampere equals 1 Coulomb per second. Since each electron has a negative charge of e≈1.602⋅10−19     C     $e\approx 1.602\cdot {10}^{-19}C$, the number of electrons per second for 1C is 1Ce≈6.242*1018 $\frac{1C}{e}\approx \mathrm{6.242*}{10}^{18}$.

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◾️Q2. Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of energy consumed in carrying a charge of 1 coulomb through a battery of 3 V.

◻️ANSWER  If 1 joule of work is done to carry 1 coulomb of charge from one point to another point , then it is said to be 1 V

Since we know that V=W/Q

so, put V = 3V And Q=1 c

by equating this in formula we get 3 j

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◾️Q3. How is an ammeter connected in a circuit to measure current flowing through it?

◽️ANSWER: The ammeter is connected ‘in series’. This means that it is connected such that all the current must flow through it.

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◾️Q4. A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new Situation.

◻️ANSWER: Given resistance = R1 = 20 ohm

Let length of the wire = L

Therefore, new length of the wire = 2L

Since, when the length is doubled, cross sectional area of the wire becomes half.

Therefore, new resistance = R = 2L/ (A/2) =4L/A

R = 4 x 20 (since, L/A = resistance)

R = 80 ohm.