Saved Bookmarks
| 1. |
A heater is designed to operate with a power of 1000 watt in a 100 V line. It is connected in combination with a resistance of 10Omega and a resistance R. to a 100 V mains as shown in the figure. What will be the value of R so that the heater operates with a power of 62.5 W? |
|
Answer» Solution :Let R. be the RESISTANCE of the heater of coil. `R=(V^(2))/(P)=((100)^(2))/(1000)=10Omega` If the heater has to operate with a power `P=62.5W`, the voltage V. ACROSS its coil should be `V.=(P.R.)^(1//2)=(6.25xx10)^(1//2)=25V` Thus, out of 100 V, a voltage DROP of 25 V occurs across the heater and the rest `100-25=75V` occurs across the `10Omega` resistor. Therefore, current in the circuit is `I=(75)/(10)=7.5A` Now, current through the heater `=(V.)/(R.)=(25)/(10)=2.5A` Therefore, current through resistor `R=7.5-2.5=5.0A` `"Hence, "R=(V.)/(5.0A)=(25V)/(5.0A)=5Omega` |
|