Board question (CBSE 2024 Main, 31/5/1 Q26): An electric source can supply a charge of 500 coulomb. If the current drawn by a device is 25 mA, find the time in which the electric source will be discharged completely.
Full Electricity long video (every repeated question type, 2022–26):
500 coulombs at 25 milliamperes: done in 20 seconds? No. Convert mA to A first, and the time is 20000 s. Solve the board question the way the marking scheme wants it: Q = It, then t = Q/I with units, then the answer with its unit.
Question (CBSE board question, 2 marks): An electric source can supply a charge of 500 coulomb. If the current drawn by a device is 25 mA, find the time in which the electric source will be discharged completely.
Given: Q = 500 C and I = 25 mA = 25 × 10⁻³ A; find t.
I = Q/t → Q = It → t = Q/I = 500 C ÷ (25 × 10⁻³ A) = 20 × 10³ s = 20000 s.
Marking scheme value points (CBSE 2024 Main 31/5/1 Q26, 2 marks): Q = It (½) · t = 500 C ÷ (25/1000 A) (½) · t = 20000 s (1).
The trap: leave the current in milliamperes and you get 20 s, which is 1000 times too small. Milli means one thousandth: 1 mA = 10⁻³ A.
Memory hook: milli means divide by a thousand.
Repeats: this charge-and-time sum (find the time or the charge with Q = It) came in 3 CBSE board papers, all 2024 Main: 31/5/1 Q26 (500 C at 25 mA, find the time), 31/5/2 Q26 (750 C at 15 mA, find the time) and 31/5/3 Q26 (0.5 A for 2 hours, find the charge). I = Q/t is also one step of a heat sum in 2025 Compartment 31/S/3 Q31(b) (72000 C in 2 hours, so I = 10 A first).
Comment challenge (CBSE 2024, assertion–reason): Assertion: electrons move from lower potential to higher potential in a conductor. Reason: a dry cell maintains electric potential difference across the ends of a conductor. Which option is right? (A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true. Answer in the pinned comment. This assertion–reason was set in 3 CBSE board papers (2024 Main 31/5/1 Q19, 31/5/2 Q20, 31/5/3 Q17).
Quick recall (Electricity, resistors in parallel): Two 4 Ω in parallel? → Rₚ = 2 Ω
Electricity quick revision (full video):
References: NCERT Class 10 Science, Ch 11 Electricity (electric current I = Q/t, the ampere and the milliampere, 1 mA = 10⁻³ A; resistors in parallel); CBSE Class 10 board questions and marking scheme.
Narration uses a licensed AI voice (Azure AI Speech, Indian English). Script and physics research by ScoreSpark.
#class10 #cbse #electricity #current #shorts
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Board question (CBSE 2024 Main, 31/5/1 Q26): An electric source can supply a charge of 500 coulomb. If the current drawn by a device is 25 mA, find the time in which the electric source will be discharged completely.
Full Electricity long video (every repeated question type, 2022–26): https://www.youtube.com/watch?v=MEab_ZNl5Lg
Options: (A) both true, R explains A · (B) both true, R does not explain A · (C) A true, R false · (D) A false, R true.
Answer: (B) Both A and R are true, but R is not the correct explanation of A. Electrons carry negative charge, so in a conductor they move from lower potential to higher potential (A true). A dry cell does maintain a potential difference across the conductor (R true), but that does not explain the direction electrons move. CBSE 2024 Main 31/5/1 Q19, 1 mark.
Board answer (2 marks, CBSE marking scheme): Q = It ½ · t = 500 C ÷ (25/1000 A) ½ · t = 20000 s 1.
Trap: 25 mA = 25 × 10⁻³ A, not 25 A (that gives 20 s). Memory hook: milli means divide by a thousand. This charge-and-time sum came in 3 CBSE board papers (2024 Main 31/5/1, 31/5/2, 31/5/3 Q26); I = Q/t is also the first step of 2025 Compartment 31/S/3 Q31(b).
Quick recall (Electricity, resistors in parallel): Two 4 Ω in parallel? → Rₚ = 2 Ω
Electricity quick revision (full video): https://www.youtube.com/watch?v=ppYotzJBczM