# AC Circuits Voltage divider rule

The basic format for the voltage divider rule in ac circuits is exactly the same as that for dc circuits: $$\bbox[10px,border:1px solid grey]{ V_x = {Z_x \over Z_T}E} \tag{1}$$ where $V_x$ is the voltage across one or more elements in series that have total impedance $Z_x$, E is the total voltage appearing across the series circuit, and $Z_T$ is the total impedance of the series circuit.
Example 1: Using the voltage divider rule, find the voltage across each element of the circuit of Fig. 1.
Fig. 1: Example 1.
Solution: $$\begin{split} V_C &= {Z_C E \over Z_C + Z_R}\\ &={(4Ω \angle -90^\circ)(100V \angle 0^\circ) \over (4Ω \angle -90^\circ) + (3Ω \angle 0^\circ)}\\ &={400 \angle -90^\circ \over 3 - j4}\\ &={400 \angle -90^\circ \over 5 \angle -53.13^\circ}=80V \angle -36.87^\circ\\ V_R &= {Z_R E \over Z_C + Z_R}\\ &={(3Ω \angle 0^\circ)(100V \angle 0^\circ) \over (4Ω \angle -90^\circ) + (3Ω \angle 0^\circ)}\\ &={300 \angle 0^\circ \over 5 \angle -53.13}\\ &=60V \angle 53.13^\circ\\ \end{split}$$
Example 2: For the circuit of Fig. 2:
a. Calculate $I$, $V_R$, $V_L$, and $V_C$ in phasor form.
b. Calculate the total power factor.
c. Calculate the average power delivered to the circuit.
d. Draw the phasor diagram.
e. Obtain the phasor sum of $V_R$, $V_L$, and $V_C$, and show that it equals the input voltage E.
f. Find $V_R$ and $V_C$ using the voltage divider rule.
Fig. 2: For Example 2.
Solution:
a. Combining common elements and finding the reactance of the inductor and capacitor, we obtain $$\begin{split} R_T &= 6 Ω + 4 Ω = 10 Ω\\ L_T &= 0.05 H + 0.05 H = 0.1 H\\ C_T &={200 \mu F \over 2} = 100 \mu F\\ \text{Now}\\ X_L &= w L_T = (377)(0.1) = 37.70 Ω\\ X_C &= {1 \over w C} = { 1 \over (377)(100 \mu F)}\\ &= 26.53 Ω\\ \end{split}$$ Redrawing the circuit using phasor notation results in Fig. 3.
Fig. 3: Applying phasor notation to the circuit of Fig. 2
For the circuit of Fig. 3, $$\begin{split} Z_T &= R \angle 0^\circ + X_L \angle 90^\circ + X_C \angle -90^\circ\\ &= 10Ω + j 37.70 Ω - j 26.53 Ω\\ &= 10Ω + j 11.17 Ω=15 \angle 48.16^\circ \\ \end{split}$$ The current I is $$I = {E \over Z_T} ={ 20 \angle 0^\circ \over 15 \angle 48.16^\circ }\\ = 1.33 A \angle -48.16^\circ$$ The voltage across the resistor, inductor, and capacitor can be found using Ohm's law: $$\begin{split} V_R &= I Z_R = (I \angle \theta^\circ)(R \angle 0^\circ)\\ &= (1.33 A \angle -48.16^\circ)(10Ω \angle 0^\circ)\\ &= 13.30 A \angle -48.16^\circ\\ V_L &= I Z_L = (I \angle \theta^\circ)(X_L \angle 90^\circ)\\ &= (1.33 A \angle -48.16^\circ)(37.70 Ω \angle 90^\circ)\\ &= 50.14 A \angle 41.84^\circ\\ V_C &= I Z_C = (I \angle \theta^\circ)(X_C \angle -90^\circ)\\ &= (1.33 A \angle -48.16^\circ)(26.53 Ω \angle -90^\circ)\\ &= 35.28 A \angle -138.16^\circ\\ \end{split}$$ b. The total power factor, determined by the angle between the applied voltage E and the resulting current I, is $48.16^\circ$: $$Fp = \cos \theta \\ = \cos 48.16^\circ = 0.667 \, lagging$$ c. The total power in watts delivered to the circuit is $$P_T = EI \cos \theta \\ = (20 V)(1.33 A)(0.667) = 17.74 W$$
d. The phasor diagram appears in Fig. 4.
Fig. 4: Phasor diagram for the circuit of Fig. 2.
e. The phasor sum of $V_R$, $V_L$, and $V_C$ is
$$\begin{split} E &= V_R + V_L + V_C \\ &= 13.30 V \angle -48.16^\circ + 50.14 V \angle 41.84^\circ + 35.28 V \angle -138.16^\circ\\ &= 13.30 V \angle -48.16^\circ + 14.86 V \angle 41.84^\circ\\ \end{split}$$
Therefore, $$E = \sqrt{(13.30V)^2+(14.86V)^2} = 20 V$$ and $\theta_E = 0^\circ$ (from phasor diagram)
and $E = 20 \angle 0^\circ$
f. $$\begin{split} V_R &= {E Z_R \over Z_T}\\ &={(10 Ω\angle 0^\circ )(20 V \angle0^\circ) \over 15Ω \angle 48.16^\circ}\\ &={200 V \angle 0^\circ \over 15Ω \angle 48.16^\circ}\\ &= 13.3 V \angle -48.16^\circ\\ V_C &= {Z_C E \over Z_T}\\ &={(26 Ω\angle -90^\circ )(20 V \angle 0^\circ) \over 15Ω \angle 48.16^\circ}\\ &={530.6V \angle -90^\circ \over 15Ω \angle 48.16^\circ}\\ &= 35.37 V \angle -138.16^\circ\\ \end{split}$$