⚑ Chapter 11: Electricity

🎯 Class 10 Science Notes (Simple + Exam Focused)


πŸ”Œ 1. Electric Current

Current (I) = Flow of electric charge
Formula:

I=QtI = \frac{Q}{t}I=tQ​

Where,

πŸ‘‰ Hindi: Vidyut dhaara charge ke flow hone ko kehte hain.


πŸ”‹ 2. Electric Potential and Potential Difference

Electric Potential: Work done to bring a unit charge from infinity to that point.
Potential Difference (V):

V=WQV = \frac{W}{Q}V=QW​

Where,

πŸ‘‰ Hindi: Do binduon ke beech ki energy difference ko potential difference kehte hain.


🧲 3. Ohm’s Law

V=IRV = IRV=IR

Where,

πŸ‘‰ Hindi: Ohm ka niyam kehta hai V current ke proportional hota hai.


🧱 4. Resistance

It is the opposition to the flow of current.
Unit: Ohm (Ξ©)
Factors affecting resistance:

πŸ‘‰ Hindi: Resistance current ke flow mein rukawat hoti hai.


πŸ”„ 5. Resistors in Series and Parallel

πŸ”Ή Series Combination:

πŸ”Ή Parallel Combination:


\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} ]

πŸ‘‰ Hindi: Series mein resistance jodte hain, parallel mein ulte jodte hain.


πŸ”₯ 6. Heating Effect of Electric Current

Formula:

H=I2RtH = I^2 R tH=I2Rt

Used in: Electric heater, bulb, geyser etc.
πŸ‘‰ Hindi: Current se heat generate hoti hai jise hum kaam mein lete hain.


πŸ”Œ 7. Electric Power

P=VI=I2R=V2RP = VI = I^2R = \frac{V^2}{R}P=VI=I2R=RV2​

πŸ‘‰ Hindi: Power ka matlab hai kaam karne ki rate.


πŸ“˜ Important Questions – With Answers


βœ… 1 Mark Questions


βœ… 2 Mark Questions

Series

Parallel

Current same everywhere

Current divides

Resistance increases

Resistance decreases

I=Qt=510=0.5 AI = \frac{Q}{t} = \frac{5}{10} = 0.5\,AI=tQ​=105​=0.5A

βœ… 3 Mark Questions

1R=12+13+16=3+2+16=1β‡’R=1 Ω\frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Rightarrow R = 1\,\OmegaR1​=21​+31​+61​=63+2+1​=1β‡’R=1Ξ©

P=VIβ‡’I=PV=100220β‰ˆ0.45 AP = VI \Rightarrow I = \frac{P}{V} = \frac{100}{220} \approx 0.45\,A P=VIβ‡’I=VP​=220100β€‹β‰ˆ0.45A R=VI=2200.45β‰ˆ489 ΩR = \frac{V}{I} = \frac{220}{0.45} \approx 489\,\OmegaR=IV​=0.45220β€‹β‰ˆ489Ξ©

βœ… 5 Mark Questions

W=VIt⇒H=I2RtW = VIt \Rightarrow H = I^2RtW=VIt⇒H=I2Rt

Use Ohm’s law to derive. Explain applications like heaters and bulbs.