# Current Electricity Key

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1. They have high resistively and low temperature coefficient of resistance.

2. 22*102 Ω ± 10%

3. Resistively remains the same.

4. As Vd α V. The different velocity will be doubled.

5. R=ρ(l/A)
= ρ(l2/Al)
= ρl2/V)
since, ρ and V are constants therefore, R α l2
 (R2/R1) = (l2/l1)2=9 because R2=9R1
=9*10=90 Ω

6. Now, 1/R=1/R1+1/R2 because l = 48/240 = 0.2 m

7. a) in parallel, power dissipation α 1/R
Therefore 3Ω wire will dissipate more power

b) In series , power dissipation α R
Therefore 9Ω wire will dissipate more power

8. R100/R27.5 = (1+100α)/(1+27.5α)
On solving, we get
α = 0.0039/˚c

9. Superconductors are the materials that lose all its resistance at very low temperature
=0 K Application:
Super conductor are used
a) In making very strong electromagnets
b) In producing very high speed computers .

10. Resistivity of copper is less , hence manganin wire is thicker.

11. High value of resistivity and low value of temperature coefficient.

12. Resistivity will be unchanged because it depends upon nature of the materials.

13. B is more sensitive.

14. Reduced by half.

15. Relaxation time decreases with increase of temperature.

16. Increase in heat.

17. Reistance remains same.

18. Pα 1/R. (i.e) 25Watts

19. 16 times of the original reisitance.

20. (i)Series - Iron
(ii)Parallel - Copper.

21. R = r L/A.
(i.e) 10.25%

22. Parallel(i.e) R.

23. 1:4.

24. 2:1

25. R.

CBSE Physics (Chapter Wise With Hint / Solution) Class XII (By Mr. Sreekumaran Nair)