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A heater coil is rated 100W,200V. It is cut into two identical parts. Both parts are connected together in parallel, to the same sources of 200V. Calculate the energy liberated per second in the new combination. |
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Answer» `because P= =(V^(2))/(R) therefore R=(V^(2))/(P)=((200)^(2))/(100)=400Omega` Resistance of half piece `=(400)/(2)=200Omega` Resistance of pieces connected in parallel `=(200)/(2)=100Omega` Energy liberated/sec ond `=P=(V^(2))/(R)=(200xx200)/(100)=400W` |
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