Electricity Class 10 Numericals
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- Опубліковано 8 вер 2024
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Electricity Class 10 Numericals : Let's solve Electricity Numericals! We will look at sums based on electric circuits, resistance, resistivity, electric power and electrical energy!
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Me also
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All Formulas of Chapter Electricity
1) I=Q/t
2) Q=ne
3) V=W/Q
4) V=IR [ V=Iyer (tmkoc wala) ]
5) R=Rho×L/A
6) H=V×Q
i) H=VIt
ii) H=I^2Rt
iii) H=V^2t/R
7) P=W/t
i) P=VI
ii) P=I^2R
iii) P=V^2/R
8) E=P×t
9) Resistance in Series-Rs=R1+R2+R3
10) Resistance in Parallel-1/Rp=1/R1+1/R2+1/R3 (i.e. Reciprocal of R1+R2+R3)
S.I. Unit
Current(I)-Ampere
Charge(Q)-Columb
Time(t)-Second
Potential Difference or Voltage(V)-Volt
Resistance(R)-Ohm
Resistivity(Rho)-Ohm meter
Power(P)-Watt
Heat=Energy=Work-Joule
Things to remember for numericals-
1 Kw-1000W
1 Horse Power-746W
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Oh my god what talent
THANKS BROO
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I'm wondering why my teacher in Physics can't explain our lessons very well not even half as good as you do.
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Sir how did you attach ammeter in parallel??
It's not possible, right?
It is in series'combination with 3 ohm
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It had been difficult for me to do the numerical now I can do with great help. So thank you for the explanation.
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10TH BOARDS KA JALWA
Sir but if we had connected the resistors in parallel then the voltage across all the resistors is equal to the voltage across battery. So we can use the potential difference of the battery as p.D of all resistors in parallel.
Note that there's one resistor which is connected in series which means it will main a different potential difference across it!
@@shrutitripathi8143 thanks. Now I understand why the potential difference across the parallel resistors will not be the same as the P.D across battery. Two resistors are in parallel. But there is one resistor is in series. Overall if we compare the individual series resisor and combination of resistors in parallel then we can easily found that the two combination are actually in series .so first we need to find voltage across the point from where the two resistors are connected in parallel .For this we need to find the equivalent resistance of the circuit which is actually 3 ohm. So the current across the whole circuit is 4A. Now it is possible to replace the combination of resistors in parallel in to single resistor of resistance 2 ohm . Now since the new resistor will be in series so same current of 4 A will pass through it and the resistance was 2 ohm .so Voltage across it will be 8 v. Now since we got the voltage across the point from where the two resistors were connected we can easily found the P.D across the R1 which is 8v .so current will be 8/3 =2.7A
You are world best teacher
Thanks for liking
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mate thats satisfying, keep it up :)
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Very Great explanation Sir.... Now I understood clearly... Thankuhh V. Much Sir
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2023-2024 wali batch ke students bhi ye video dekh rahe hai pre-board ke liye 😅
Sir please make a seperate video on the difficult numericals based on the connection of resistors in the circuit
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SIR ONE MORE VIDEO ON NUMERICAL OF ELECTRICITY.
PLEASE SIR. YOUR EXPLAINATION IS VERY HELPFUL FOR ALL CLASS 10 STUDENT.
Plz put important numeicals in magnetic effects of current.
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But the voltage is the same through all the resistors in a parallel combination.Then why can't we use 12v directly?
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Thank goodness I found a physic teacher who made physic numericals easy to me
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O u can't sleep 😁😂
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Glad you liked it
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For the first question. Can't we do it like 1/Rp = 3ohm + 1ohm (series connection) hence I = V/R 12 /4 = 3 ohm (approximately)
Sir in 1st Q you took the V1=8V for the 3 Ohms resistor Connected in parallel Comb. but I've read that voltage will remain same in parallel resistors, then how it'll change Sir?
Dear sir,your teaching methods are excellent which help me and learners a lots.Pl,keep it up ! Thanks!
Thank you, I will