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a. For circuits used for transporting electric power, a low power factor implies large powerloss in transmission. Explain.b. Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. |
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Answer» SOLUTION :a: We know that `P= IV cos phi`where `cos phi`is the power factor. To supply a given power at agiven voltage, if cos `phi`is small, we have to increase current accordingly. But this will lead to large power loss (`I^2R` ) in transmission. b. Let the current I lags the voltage by an angle. Then power factor` cos phi = R/Z` ` cos phi` can be improved (say tending to I) by making Z tend to R. Let us understand, with the help of a phasor DIAGRAM (Figure) how this can be achieved. Let us resolve I into two components. I along the applied voltage V and `I_q`perpendicular to the applied voltage. `I_q`I is called the wattless component since corresponding to this component of current, there is no power loss. `I_p`is KNOWN as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It.s clear from this analysis that if we want to improve power factor we must completely neutralize the lagging wattless current `I_q`by an EQUAL leading wattless current `I_q` . This can be DONE by connecting a capacitor of appropriate value in parallel so that `I_q`and `I._q`cancel each other and P iseffectively `I_p V` 
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