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Biology; Chemistry; Physics: Post your doubts here!

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Its A. but why?
oh yeah we had to consider the conventional current -_-

it should be A cuz diodes make sure that the current passes in the right direction and if it is placed on the negative line (b) then the current would stop making it a reverse biased junction....LOL
not sure of this :unsure:
 
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erm didn't get you.
well it was just a try.... i thought of something... ok take it in a way like the movement of ions are in electrolysis... the diode is taking the elecrtons from the positive side (as electricity passes from +ve side to -ve side) to the battery from where the ellectrons go to the negative and the current is flowed... I dnt think i em ryt but i wanted to give a try/../....
 

asd

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Its A. but why?
In A the current can flow as long as the battery connections are reversed. You see the Pointed part of the diode indicates the direction where it allows the current to flow. So if you reverse the connections to the battery, the current will not flow, because then the pointed part of the diode will be opposing the current flow!
 
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In A the current can flow as long as the battery connections are reversed. You see the Pointed part of the diode indicates the direction where it allows the current to flow. So if you reverse the connections to the battery, the current will not flow, because then the pointed part of the diode will be opposing the current flow!

well it was just a try.... i thought of something... ok take it in a way like the movement of ions are in electrolysis... the diode is taking the elecrtons from the positive side (as electricity passes from +ve side to -ve side) to the battery from where the ellectrons go to the negative and the current is flowed... I dnt think i em ryt but i wanted to give a try/../....

Thanks alot guys!
 
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wv2j.jpg


Why is the friction force acting towards D instead of B??
 

asd

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Well all other options seem idiotic, i go with B, because more surface area, more electrical field, and therefore more deflection.
Decreasing the distance will DECREASE the deflection, since the spot will be on the screen even when it hasn't deflected fully.
Increasing the potential b/w cathode and anode has no relation to the deflection. Only the potential provided to the plates matters here.
Increasing the separation will make the electric field weaker! so LESS deflection.
 
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