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Suppose the circuit in Exercise 18 has a resistance of `15 Omega`. Obtain the average power transferred to each element of the circuit, and the total power absorbed. |
Answer» Given, the rms value of voltage `V_("rms")=230 V` Resistance `R = 15 Omega` Frequency `f = 50 Hz` Average power across inductor and capacitor is zero as the phase difference between current and voltage is `90^(@)`. Total power absorbed = power absorbed in resistor, `P_("av")` `= V_("rms").I_("rms")` So, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))` `= sqrt((15)^(2)+(2xx3.14xx50xx80xx10^(-13)-(1)/(2xx3.14xx50xx60xx10^(-6))))` `= sqrt(1002.85)=31.67 Omega` `I_("rms")=(V_("rms"))/(Z)=(230)/(31.67)=7.26 A` Total power absorbed `P_("av")` `= V_("rms").I_("rms")` `= I_("rms").R.I_("rms")` `= (7.26)^(2)xx15=790.6W` Total power absorbed = 790.6W |
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