plz sir next chapter - light aur sb ka ho jata h mei pw ka batch liya hue lekin abhi tak light ka backlog nhi khatam hua h isseleye sir plz next chapter -light kerwa dejeye ok aapka pyara warrior student❤❤❤🎉
All Formulas of Electricity: 1) I = Q/t 2) Q = ne 3) V = W/Q 4) V = IR 5) R = Rho x L/A 6)Resistance in Series: Rs = R1 + R2 + R3 7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3) 8)H = V × Q i) H = VIt ii) H = I²Rt iii) H = V²t/R 9) P = W/t i) P = VI ii) P = I²R iii) P = V²/R 10) E = P × t S.I. Units: Current ( I ) - Ampere Charge ( Q ) - Coulomb Time ( t ) - Second Potential Difference or Voltage ( V ) - Volt Resistance ( R ) - Ohm Ω Resistivity ( Rho ) - Ohm meter Power ( P ) - Watt Heat = Energy = Work = Joule Measuring Devices: Ammeter / Milli Ammeter - for measuring current (always connected in series) Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current) Voltmeter - for measuring volts (always connected in parallel) Ohm Meter - Measures the resistivity Things to remember: electron charge = (1.6 x 10⁻¹⁹ C) 1kWh= 3.6 × 10⁶J 1Ampere = 1000mA 1 Kilowaat = 1000W 1 Horse Power = 746W i hope this list will help you! ❤ ek free ka like dete jana 👇 Edit : OMG.. I was not expecting soo many likes. Thank you all..!!!!🤩
1) I = Q/t 2) Q = ne 3) V = W/Q 4) V = IR 5) R = Rho x L/A 6)Resistance in Series: Rs = R1 + R2 + R3 7) Resistance in Parallel: 1/Rp = 1/R1 + 1/ R2 + 1/R3 (» Reciprocal of R1+R2+R3) Energy/Heat = VQ i) H = VIt ii) H = I ^ 2 * Rt iii) H = V^2/R×t 9) P = W/t i) P= VI ii) P = I^2 * R iii) P = (V^2) / R 10) E=P×t S.I. Units: Current (I)- Ampere Charge (Q) Coulomb Time (t)- Second Potential Difference or Voltage (V) - Volt Resistance (R) Ohm Ω Resistivity (Rho) - Ohm meter Power (P) - Watt Heat = Energy = Work = Joule Measuring Devices: Ammeter / Milli Ammeter for measuring current (always connected in series) Galvanometer for measuring small/sensitive currents (we can also get the direction of the current) Voltmeter for measuring volts (always connected in parallel) Ohm Meter - Measures the resistivity Things to remember: electron charge = (1.6 * 10 ^ - 19 * C) 1kWh = 3.6 * 10 ^ 6 * J 1 Ampere 1000mA 1 Kilowaat = 1000W 1 Horse Power = 746W
GUYS WAISE TO MAINA QUESTION ROUGH SOLVE KARA THA PER IMPERSION KA LIYA MAINA STEP BY STEP LIKHA HA😅😅😅 QUESTION 2:26:54 Solution Step 1: Identify the circuit type. The circuit is a combination of series and parallel circuits. Step 2: Calculate the equivalent resistance. The 3Ω resistors in parallel have an equivalent resistance of 1.5Ω. The equivalent resistance of the entire circuit is then 1.5Ω + 3Ω + 3Ω = 7.5Ω. Step 3: Calculate the total current. Using Ohm's Law (I = V/R), the total current in the circuit is 10V / 7.5Ω = 1.33A. Step 4: Calculate the current through A. The current through A is the same as the total current, which is 1.33A. Step 5: Calculate the current through A1. The current splits at the junction where the 3Ω resistor connects to the parallel resistors. Since the parallel resistors have equal resistance, the current splits equally. Therefore, the current through A1 is 1.33A / 2 = 0.665A. Answers A = 1.33A A1 = 0.665A
One of the best faculty of physics Rakshak sir ❤ 9th class me I started studying from his yt lectures,I hated physics but after studying from him I loved physics and till date I love physics
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@@PW-Foundation pls sir next chapter - light . baaki easy hai .
plz sir next chapter - light aur sb ka ho jata h mei pw ka batch liya hue lekin abhi tak light ka backlog nhi khatam hua h isseleye sir plz next chapter -light kerwa dejeye ok aapka pyara warrior student❤❤❤🎉
Sir light krwa do
Yess sir light so time taking chapter
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All Formulas of Electricity:
1) I = Q/t
2) Q = ne
3) V = W/Q
4) V = IR
5) R = Rho x L/A
6)Resistance in Series: Rs = R1 + R2 + R3
7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3)
8)H = V × Q
i) H = VIt
ii) H = I²Rt
iii) H = V²t/R
9) P = W/t
i) P = VI
ii) P = I²R
iii) P = V²/R
10) E = P × t
S.I. Units:
Current ( I ) - Ampere
Charge ( Q ) - Coulomb
Time ( t ) - Second
Potential Difference or Voltage ( V ) - Volt
Resistance ( R ) - Ohm Ω
Resistivity ( Rho ) - Ohm meter
Power ( P ) - Watt
Heat = Energy = Work = Joule
Measuring Devices:
Ammeter / Milli Ammeter - for measuring current (always connected in series)
Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current)
Voltmeter - for measuring volts (always connected in parallel)
Ohm Meter - Measures the resistivity
Things to remember:
electron charge = (1.6 x 10⁻¹⁹ C)
1kWh= 3.6 × 10⁶J
1Ampere = 1000mA
1 Kilowaat = 1000W
1 Horse Power = 746W
i hope this list will help you! ❤
ek free ka like dete jana
👇
Edit : OMG.. I was not expecting soo many likes. Thank you all..!!!!🤩
Thanks bro
thnx
Thanks Bhai
Thanks you 😊
Phhle se hi edit krr diya comment wahhh😂
Magnetic Effects of Electric Current next!!!!!
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2:26:48 Answer is - A = 5A & A1 = 3.333A
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1) I = Q/t
2) Q = ne
3) V = W/Q
4) V = IR
5) R = Rho x L/A
6)Resistance in Series: Rs = R1 + R2 + R3
7) Resistance in Parallel: 1/Rp = 1/R1 + 1/
R2 + 1/R3 (» Reciprocal of R1+R2+R3)
Energy/Heat = VQ
i) H = VIt
ii) H = I ^ 2 * Rt
iii) H = V^2/R×t
9) P = W/t
i) P= VI
ii) P = I^2 * R
iii) P = (V^2) / R
10) E=P×t
S.I. Units:
Current (I)- Ampere
Charge (Q) Coulomb
Time (t)- Second
Potential Difference or Voltage (V) - Volt
Resistance (R) Ohm Ω
Resistivity (Rho) - Ohm meter
Power (P) - Watt
Heat = Energy = Work = Joule
Measuring Devices:
Ammeter / Milli Ammeter for measuring current (always connected in series)
Galvanometer for measuring small/sensitive
currents (we can also get the direction of the current)
Voltmeter for measuring volts (always connected in parallel)
Ohm Meter - Measures the resistivity
Things to remember:
electron charge = (1.6 * 10 ^ - 19 * C)
1kWh = 3.6 * 10 ^ 6 * J
1 Ampere 1000mA
1 Kilowaat = 1000W
1 Horse Power = 746W
Sir please 🥺
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2:26:45
A=5A
A1=3.33A
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2:26:57 homework.
A = 5 A
A1=10/3A.❤❤
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2:26:48 HOMEWORK QUESTION
Reading in ammeters A and A1 will be 5A and 3.3A respectively
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GUYS WAISE TO MAINA QUESTION ROUGH SOLVE KARA THA PER IMPERSION KA LIYA MAINA STEP BY STEP LIKHA HA😅😅😅
QUESTION 2:26:54
Solution
Step 1: Identify the circuit type. The circuit is a combination of series and parallel circuits.
Step 2: Calculate the equivalent resistance. The 3Ω resistors in parallel have an equivalent resistance of 1.5Ω. The equivalent resistance of the entire circuit is then 1.5Ω + 3Ω + 3Ω = 7.5Ω.
Step 3: Calculate the total current. Using Ohm's Law (I = V/R), the total current in the circuit is 10V / 7.5Ω = 1.33A.
Step 4: Calculate the current through A. The current through A is the same as the total current, which is 1.33A.
Step 5: Calculate the current through A1. The current splits at the junction where the 3Ω resistor connects to the parallel resistors. Since the parallel resistors have equal resistance, the current splits equally. Therefore, the current through A1 is 1.33A / 2 = 0.665A.
Answers
A = 1.33A
A1 = 0.665A
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A-5a/A1-3.333
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2:26:15
A = 5A
A¹ = 3.33 A
A=5ohm
A1=3.33ohm 2:26:56
Bro, not ohm sI unit is ampere (A)
Don't forget this
H.w
A=5A
A1 = 3.33A
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A=5A
A1=3.33A
A-5a/A1-3.333a
2:26:48 Answer A=5A
A1=10/3A or 3.333...A
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Answer
A=5
A1=3.34
Homework Question:-
5Ampere
3.33Ampere
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1:03:02
Ritik sir ❤❤❤❤❤😂😂
2:26:42
Sir ammeter A shows 5 ampere of current
And A1 shows 3.33 ampere
A=5A
A1=3.33 A
🎉🎉🎉 1:29 1:29
A- 5Ampere
A1- 3.33 A
3:03:28 thank you sir
Ampere is 5 and ampere 1 is 10 by 3 2:26:09
Ok
Ok
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Ans, A=5A
A1 =3.33A
Sir thankyou for your humanity
Honework question
A=5A
A1=3.33 Aor 3.4 A (approx)
A-5A A1-3.33
Can you helpout
Actually sir has told us that voltage will divide among 3 ohm resistor so shouldn't we minus it?
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HOMEWORK A= 5A & 2ND ANS IS A1=1.66
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One of the best faculty of physics Rakshak sir ❤ 9th class me I started studying from his yt lectures,I hated physics but after studying from him I loved physics and till date I love physics
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A=5A
A¹=3.3A
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