Answer: The SI unit of electric current is ampere (A) . 1 ampere = 1 coulomb per second (1 A = 1 C/s).
Answer: Total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω ( I = 9 / 13.4 \approx 0.67 , \text{A} ) (same in series)
Answer: (i) 9 Ω = two in parallel (3 Ω) + one in series (6 Ω) (ii) 4 Ω = two in series (12 Ω) in parallel with one (6 Ω): ( 1/R = 1/12 + 1/6 = 3/12 ) → R=4 Ω
Answer: From Ohm’s law ( V = IR ), if ( R ) constant and ( V ) becomes half, ( I ) also becomes half.
Answer: Alloys have higher resistivity and do not oxidize easily at high temperatures.
Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below: I (amperes) : 0.5, 1.0, 2.0, 3.0, 4.0 V (volts) : 1.6, 3.4, 6.7, 10.2, 13.2 Plot V–I graph and calculate resistance. Slope ( \Delta V / \Delta I ) ≈ 3.3 Ω.
Class 10 Electricity Ncert Solutions Best May 2026
Answer: The SI unit of electric current is ampere (A) . 1 ampere = 1 coulomb per second (1 A = 1 C/s).
Answer: Total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω ( I = 9 / 13.4 \approx 0.67 , \text{A} ) (same in series) class 10 electricity ncert solutions
Answer: (i) 9 Ω = two in parallel (3 Ω) + one in series (6 Ω) (ii) 4 Ω = two in series (12 Ω) in parallel with one (6 Ω): ( 1/R = 1/12 + 1/6 = 3/12 ) → R=4 Ω Answer: The SI unit of electric current is ampere (A)
Answer: From Ohm’s law ( V = IR ), if ( R ) constant and ( V ) becomes half, ( I ) also becomes half. Answer: Alloys have higher resistivity and do not
Answer: Alloys have higher resistivity and do not oxidize easily at high temperatures.
Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below: I (amperes) : 0.5, 1.0, 2.0, 3.0, 4.0 V (volts) : 1.6, 3.4, 6.7, 10.2, 13.2 Plot V–I graph and calculate resistance. Slope ( \Delta V / \Delta I ) ≈ 3.3 Ω.