As shown in Fig. 35-3, a series circuit connected across a 200-V, 60 - Hz line consists of a cap

step 1 We have to find the value of power potential difference across the capacitor register and power

As shown in Fig. 35-3, a series circuit connected across a...

As shown in Fig. $35-3$, a series circuit connected across a $200-V$, 60 - $\mathrm{Hz}$ line consists of a capacitor of capacitive reactance $30 \Omega, \mathrm{a}$ noninductive resistor of $44 \Omega$, and a coil of inductive reactance 90 $\Omega$ and resistance $36 \Omega$. Determine $(a)$ the current in the circuit, $(b)$ the potential difference across each element, (c) the power factor of the circuit, and $(d)$ the power absorbed by the circuit. Fig. $35-3$ (a) $\quad Z=\sqrt{\left(R_{1}+R_{2}\right)^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(44+36)^{2}+(90-30)^{2}}=0.10 \mathrm{k} \Omega$ So $\quad I=\frac{V}{Z}=\frac{200 \mathrm{~V}}{100 \Omega}=2.0 \mathrm{~A}$ (b) p.d. across capacitor $=I X_{C}=(2.0 \mathrm{~A})(30 \Omega)=60 \mathrm{~V}$ p.d. across resistor $=I R_{1}=(2.0 \mathrm{~A})(44 \Omega)=88 \mathrm{~V}$ D.d. across coil $=(2.0 \mathrm{~A})(97 \Omega)=0.19 \mathrm{kV}$ (c) Powder factor $=\cos \phi=\frac{R}{Z}=\frac{80}{100}=0.80$ $(d)$ Power used $=V / \cos \phi=(200 \mathrm{~V})(2 \mathrm{~A})(0.80)$

As shown in Fig. $35-3$, a series circuit connected across a $200-V$, 60 - $\mathrm{Hz}$ line consists of a capacitor of capacitive reactance $30 \Omega, \mathrm{a}$ noninductive resistor of $44 \Omega$, and a coil of inductive reactance 90 $\Omega$ and resistance $36 \Omega$. Determine $(a)$ the current in the circuit, $(b)$ the potential difference across each element, (c) the power factor of the circuit, and $(d)$ the power absorbed by the circuit.
Fig. $35-3$
(a) $\quad Z=\sqrt{\left(R_{1}+R_{2}\right)^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(44+36)^{2}+(90-30)^{2}}=0.10 \mathrm{k} \Omega$
So $\quad I=\frac{V}{Z}=\frac{200 \mathrm{~V}}{100 \Omega}=2.0 \mathrm{~A}$
(b) p.d. across capacitor $=I X_{C}=(2.0 \mathrm{~A})(30 \Omega)=60 \mathrm{~V}$
p.d. across resistor $=I R_{1}=(2.0 \mathrm{~A})(44 \Omega)=88 \mathrm{~V}$
D.d. across coil $=(2.0 \mathrm{~A})(97 \Omega)=0.19 \mathrm{kV}$
(c) Powder factor $=\cos \phi=\frac{R}{Z}=\frac{80}{100}=0.80$
$(d)$ Power used $=V / \cos \phi=(200 \mathrm{~V})(2 \mathrm{~A})(0.80)$

Key Concepts

Voltage Division in Series AC Circuits

Voltage division in series AC circuits involves the distribution of the total applied voltage across individual components according to their impedance values. Each element in the series circuit, whether resistive or reactive, drops a fraction of the total voltage proportional to its impedance, following a similar principle to voltage division in purely resistive circuits but extended to include the phase differences introduced by reactance.

Power Factor

Power factor is the cosine of the phase angle between the voltage and the current in an AC circuit. It indicates the fraction of the total power that is effectively used for work. A power factor close to 1 means that most of the power is being effectively utilized, while a lower power factor indicates that a larger portion of the power is reactive and does not contribute to useful work.

Impedance in AC Circuits

Impedance is the combined effect of resistance and reactance in an AC circuit and represents the total opposition to the current flow. It is a complex quantity that is obtained by vector addition of the resistance (real part) and the net reactance (imaginary part), and it determines the overall current when the circuit is connected to an AC voltage source.

Reactance (Inductive and Capacitive)

Reactance is the opposition to the change of current in AC circuits provided by inductors and capacitors. Inductive reactance increases with frequency and tends to cause the current to lag the voltage, while capacitive reactance decreases with frequency and tends to cause the current to lead the voltage. The difference between these reactances determines the net reactive effect in a circuit.

Power Calculation in AC Circuits

AC power calculation takes into account not only the voltage and current but also the phase difference between them. The real power, which is the power actually absorbed by the circuit, is calculated by multiplying the voltage, current, and the power factor. This demonstrates the importance of both the magnitude of the current and the relative phase between the voltage and current in determining the power consumption in AC circuits.

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