Electricity Class 10 One-Shot | CBSE Class 10 (Chapter 12) Science Physics (Term-2) Crash Course

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In this Episode of Class 10 One-Shot Series, we take up CBSE Class 10 Science Chapter 12 - Electricity. We help you understand all concepts related to Ohm's Law, Resistance, Factors Affecting Resistance, Resistivity, Combination of Resistors, Advantages of Parallel Circuit Over Series Circuit, Heating Effect of Electric Current, Joule's Law of Heating, Electrical Power, Commercial Unit of Electrical Energy and much more.

In this Term-2 Crash Course Series, we will discuss the important topics of Electricity which help you to solve CBSE Important questions and get prepared for Class 10 Term-2 Board Exams.

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00:00 Introduction
3:23 Electricity
3:46 Experiment
5:58 Experiment: Inference
6:42 Ohm’s Law
7:26 Resistance (R)
8:50 Factors Affecting Resistance of a Conductor
10:36 Dependence of Resistance on Material
11:23 Conductor and Insulators
12:24 Dependence of Resistance on Temperature
13:24 Resistivity
18:23 Factors affecting R and p
19:06 Ohm’s Law
20:09 Combination of Resistors
20:48 Series Combination of Resistors
24:48 Equivalent Resistance of ‘n’ number of equal resistors in series
25:27 Parallel Combination of Resistors
26:57 Equivalent Resistance of ‘n’ number of equal resistors in Parallel
27:31 Series Vs Parallel Combination
27:38 Advantages of Parallel Combination
28:56 Combination of Resistors
29:46 Heating Effect of Electric Current
32:40 Joule’s Law of Heating
33:00 Application of Heating Effect of Electric Current
33:55 Electric Bulb
36:11 Fuse - The safety Device
38:19 Heating Element
40;10 Electric Power
41:56 Commercial Unit of Electrical Energy
43:30 Homework Question

#ByjusClass10 #CbseClass10 #Class10Science #Electricity #ElectricityClass10 #CBSETerm2 #Term2Preparations #Term2CrashCourse #Term2Science #Class10Physics #Class10Term2 #Byjus #Liveclasses #Onlineclasses #NCERTbooks

[Systemmm_hang_]

Part1.
Power = I^2 R
here I = 2A and R = 1ohm
Therefore,
(2^2)A × 1 ohm = 4A×1ohm
= 4 W
Part2.
Given that, V(potential difference) = 10Volts
I (current). = 2A
Resistance = V/A
= 10V / 2A
=5ohm
Total Resistance=5ohm
We have three resistors in the circuit.
Two of them are in parallel and the other in series
We have Rs = 1ohm and
1/Rp= 1/8 ohm + 1/R ohm
= 8+R / 8R
Rp= (8R/8+R) ohm
Total Resistance = Rp + Rs
5 ohm = (8R/8+R)ohm + 1 ohm
5 - 1 = 8R/8+R
4 = 8R/8+R
4(8+R) = 8R
32+4R = 8R
32 = 8R - 4R
32 = 4R
R = 32/4
R = 8ohm
So the answers are :- 1.Power consumed by 1ohm resistor is 4 W
2. The required value of R is 8 ohm
Thank you Byju's for such interesting questions and amazing sessions...
🥰🥰🥰

[abigaildavid]

Why does he read the chats more than teaches ? 😑

[Ashishkumar-pncq]

Homework :-
1) power consumed by 1 ohm resistor = 4 Watt
2) R = 8 ohm
😊😊😊

Solution:

1) Resistance = 1ohm
Current in circuit = 2 A
Therefore Potential difference across the resistor (1ohm) = V = RI = 2×1 = 2 V
=> power = VI = 2×2 = 4 Watt

2) Total resistance in the circuit = Potential diff. across the battery/ current flowing in the circuit = 10/2 = 5 ohm
and we have 1 ohm resistor in series, therefore equivalent resistance for the other two parallel resistor must be "4".

thus, 1/8 + 1/R = 1/4 => 1/R = 1/4 - 1/8 => 1/8
Hence R = 8

[hansikahoney]

1) Resistance = 1ohm
Current in circuit = 2 A
Therefore Potential difference across the resistor (1ohm) = V = RI = 2×1 = 2 V
=> power = VI = 2×2 = 4 Watt

2) Total resistance in the circuit = Potential diff. across the battery/ current flowing in the circuit = 10/2 = 5 ohm
and we have a 1-ohm resistor in series, therefore equivalent resistance for the other two parallel resistors must be "4".

thus, 1/8 + 1/R = 1/4 => 1/R = 1/4 - 1/8 => 1/8
Hence R = 8

[universalchoice]

3:20 video starts











Welcome 🙃

[aditidutta]

Homework answer
i. Power consumed= 4 w
ii. R=8 ohm

[charvi]

Homework:
Power consumed by the 1Ω resistor is given by I²R = (2A)²(1Ω) = 4W.
Now, the equivalent resistance of the 8Ω resistor and R = (1/8 + 1/R)^(-1) Ω= {[(R + 8)]/8R}^(-1) Ω = 8R/(R + 8) Ω
And, the equivalent resistance of the whole circuit = 8R/(R + 8) Ω + 1Ω ...(1)
Also, the equivalent resistance of the whole circuit, R' = V/I = 10V/2A = 5Ω ...(2)
From (1) and (2), we get
8R/(R + 8) Ω + 1Ω = 5Ω
8R/(R + 8) Ω = 4Ω
8R = 4R + 32
4R = 32
R = 8
Therefore, the value of R is 4Ω.

Great session. Thank you very much sir. 😊😊

[aiswarya]

Nice session ☺
Hw 1) 4 W
2) 8 ohm
Typing the steps takes time.
Thank you byjus for the amazing class.

[puneet_kaur]

Homework :1•Power consumed=4watt
2•Resistance =8‌ohm
1•R=1ohm
I=2A
V=IR=2×1=2V
P=VI=2×2=4WATT
2•V=10V
R=V/I
R=10/2=5ohm
1/Rp=1/8ohm+1/Rohm
Rp=8R/8+R
Rs+Rp=5=8R/8+R+1
=8ohm

[janyasingh]

Homework
Power consumed= 4 watts
Resistance R= 8 ohms

[kajalKumari-roye]

Homework :✨
1) power consumed by 1 ohm resistor = 4 Watt
2) R = 8 ohm
Solution:
1) Resistance = 1ohm
Current in circuit = 2 A
Therefore Potential difference across the resistor (1ohm)
=V=RI=2X1=2
=> power = VI = 2X2 = 4 watts
2) Total resistance in the circuit = Potential diff. across the battery/ current flowing in the circuit = 10/2 = 5 ohm and we have 1 ohm resistor in series, therefore equivalent resistance for the other two parallel resistor must be "4".
thus. 1/8 + 1/R = 1/4 => 1/R = 1/4 - 1/8 => 1/8
Hence R = 8

[bhagyavardhan]

Answer to homework questions
(1) Power consumed by 1 ohm resistor
R= 1 ohm., I = 2 A
V = 2A × 1 ohm = 2 V
P = VI = 2 V × 2 A = 4 W
So, power consumed by 4 ohm resistor is 4 Watt.
(2) V = 10 V. I = 2A
R = 10/2. = 5 Ohm. ,
So, 1/8 + 1/ R = 1/4
1/ R = 1/8. So, resistance across R is 8 ohm.

[nikeshmadavi]

Homework Answers :-
Equivalent resistance in circuit, R = V/I
Req = 5Ω
Applying equavalent resistance formula for series and parallel combination with unknown R and equating it with 5 Ω we get,
Value of R as 8 Ω


Power consumed by 1 Ω resistor is I²R which is 4 watt

[lavyagoyal]

Why current is different in parallel combination

[rreya_]

Homework Answer :
The power consumed by the 1 Ω resistor is 4 watts and the value of R in the given circuit is 8 Ω.

[girishsarus]

Power consumed by 1 ohm resistor is I^2R
Therefore, 2A^2×1=4W

And the total resistance is R=V/I
That is 10V/2A=5 ohms

Therefore,
1/R1+1/R2= 1/ R3
1/8+1/R=1/R3
R3=8R/R+8

Then,
(8R/R+8) +1=5 ohms
R=8 ohms

Homework completed sir😊😊

[shaikrehan]

Sir can you send me the video of mole concept

[BYJUSClass]

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[cherrycharitha-zh]

Would appreciate it greatly if the teachers didn't read the chat all the time

[inspirit..]

Thank u sir for providing us such a quality content 😊