Saved Bookmarks
| 1. |
Three equal resistors connected in series across a source of emf together dissipate 10 watt of power. What would be the power dissipated if the same resistances are connected in parallel acrsoss the same source of emf? |
|
Answer» SOLUTION :The power consumed by a RESISTANCE R when CONNECTED across a source of emf V is given by `P=(V^(2)//R)` Now, if r is the resistance of each resistor, the resistance of COMBINATION, in series will be `R_(s)=r+r+r+i.e…,"R_(S)=3r` while in paralle, `(1)/(R_(p))=(1)/(r)+(1)/(r)+(1)/(r)"i.e...,"R_(p)=(r)/(3)` So power consumption in series will be, `P_(S)=(V^(2))/(3r)["as"R_(S)=3r]` while in parallel will be, `P_(p)=(V^(2))/((r//3))=3[(V^(2))/(r)][" as "R_(p)=(r)/(3)]` `"i.e..,"(P_(p))/(P_(s))=3[(V^(2))/(r)]xx(3r)/(V^(2)]=9` And as here `P_(s)=10W` `P_(P)=9xx(P_(S))=9xx10=90W.` |
|