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An electric tea kettle has two heating coils. When one of the coils is switched on, boiling begins in 6 min. When the other coil is switched on, the boiling begins in 8 min. In what time, will the boiling begin if both coils are switched on simultaneously (i) in series and (ii) in parallel. |
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Answer» Solution :Let power of first COIL is `P_(1)` and that of SECOND coil is `P_(2)`. Let H is the amount of heat required to oil water. Then `H=P_(1)t_(1)=p_(2)t_(2)` where `t_(1)="6 MIN,"t_(2)="8 min"` i) When the coils are connected in series : `P=(P_(1)P_(2))/(P_(1)+P_(2))""t=(H)/(P)=H[(1)/(P_(1))+(1)/(P_(2))]=H[(t)/(H)+(t_(2))/(H)]` `=t_(1)+t_(2)=6+8="14 min"` ii) When the coils are connected in parallel `P=P_(1)+P_(2)` `t=(H)/(P)=(H)/(P_(1)+P_(2))=(H)/((H)/(t_(1))+(H)/(t_(2)))=(t_(1)t_(2))/(t_(1)+t_(2))=(6xx8)/(6+8)="3.43 min."` |
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