- Messages
- 46
- Reaction score
- 21
- Points
- 18
Ok no problemo.For the question no33,i think it a bit logically.Path B shows path taken by proton.So now for helium nucleus it is
4 1
2 he 1 p.
------------>the helium is much heavier than the proton.....So,it would need a greater distance to be completely deflected than
the proton..For the direction,it should be same as the proton as both r same charge.
For the second question mean power dissipated is done by calculating the= highest power+lowest power/2
Highest power=i^2*r==>2^2*100=400 watt
lowest power=-1^2*100=100watt
400+100/2=250 watt
For the third question,,
option a rite-----Because when the s1 is closed,althugh s2 is also closed,the current will instead pass the region
without resistance,path s1.So the overall circuit becomes like the original circuit above.The r at s2 have no effect onn the current reading
option b rite--i think u know why.Its exactly similar to above circuit.No current flow to the bottom R thus,same ammeter reading
option d rite-----no current flow as switch all open.No complete cicruit
Option c is WRONG.When s1 is open n S2 is closed,the current must now pass the extra resistor as it has no other chance.Thus overall resistance
is higher n current value is less than I.
Awwww thank you o(≧▽≦)o