Question

1. Using circuit 3-1, calculate the total current (which is also the capacitor current and resistor current) by using Ohm’s Law. To do this, you must first compute the total impedance of the circuit, in polar form. Also, remember that Vs (source voltage) phase shift is 0 degrees. Write your answer in polar form.

2. Compute the voltage across the capacitor (C1), using Ohm’s Law and your result from #1. Write your answer in polar form.

3. In a series RC circuit, the voltage across the capacitor lags the total current by 90o . Using your answer for #1 and #2, verify that this is true. What is the phase difference between your calculated values for IT and VC?

4. Compute the voltage across the resistor (R1), using Ohm’s Law. Write your answer in polar form.

5. In a series RC circuit, the voltage across the resistor is in phase with the current flowing through it. Using your answer for #1 and #2, verify that this is true. Are VR and IT in phase with each other?

R1 270 Ohm /1v@ 1 KHz C1 1 uF Figure 1: Circuit 3-1

Answer 1

The given circuit is a RC circuit with R = R₁ = 270 ohm and C = C₁ = 1uF.

1. To calculate the total current flowing through the circuit, we first find the total impedance.
   Let Z be the total impedance, R be the resistance and X be the reactance of the circuit.
   Note that reactance refers to the imaginary part of the impedance which is contributed by inductors and capacitors, in this case the capacitor C₁.
   For a capacitor we denote the reactance by X_C.
   $X_C = \frac{1}{2\pi fC}$
   here f represents the frequency of the signal in Hz which in this case is 1kHz as seen from the circuit diagram.
   $f = 1kHz = 10^3 Hz, C = 1\mu F = 10^-^6 F, \pi = 3.14$
   $X_C = \frac{1}{2\times3.14\times10^3\times10^-^6} \Omega$
   $X_C = 159.2\ \Omega$
   $R = 270\ \Omega$
   
   Now the overall impedance of the circuit is given by
   $Z = R-jX_C\ \Omega$
   $Z = 270-j159.2\ \Omega$
   To convert it to polar form, we find the modulus of Z and its phase in degrees
   $|Z| = $\sqrt{270^2+159.2^2}$\ \Omega$
   $|Z| = 313.4\ \Omega$
   
   Phase in degrees
   $\angle Z = \arctan({\frac{-159.2}{270}})$
   $\angle Z = -30.52^0$
   Hence the impedance in polar form is given by
   $Z =313.4\ \angle-30.52^0 \ \Omega$
   
   Voltage of the source is given as
   $V_s =1\ \angle0^0 \ V$
   
   Hence from Ohm's law the current flowing through the circuit is given by
   $I_T = \frac{V_s}{Z}$
   $I_T = \frac{1\ \angle0^0}{313.4\ \angle-30.52^0} \ A$
   $I_T =0.00319 \angle30.52^0 \ A$
   $I_T =3.19\ \angle30.52^0 \ mA$
   
   Hence the total current flowing through the circuit in polar form is given by
   $I_T =3.19\ \angle30.52^0 \ mA$
   
2. Voltage across capacitor C₁ is proportional to the current flowing through C₁.
   From part 1, we have
   
   $X_C = 159.2 \ \Omega$
   Therefore, impedance of the capacitor Z_C can be written as
   $Z_C = X_C \ \angle-90^0$
   $Z_C = 159.2 \ \angle-90^0 \ \Omega$
   Hence by Ohm's law voltage across the capacitance V_C is given as
   $V_C = I_TZ_C$
   $V_C = 3.19\ \angle30.52^0 \times 159.2\ \angle-90^0 \ mV$
   $V_C = 507.8\ \angle-59.45^0 \ mV$
   
   Therefore voltage across C₁ is
   $V_C = 507.8\ \angle-59.45^0 \ mV$
   
3. From the above values, the phases of I_T and V_C are given by
   $\angle V_C = -59.45^0$
   $\angle I_T = 30.52^0$
   
   Hence phase difference is
   $-59.45^0-30.52^0 = 89.97^0 \approx -90^0$
   
   Therefore we verified that the voltage across the capacitor V_C lags the total current I_T by 90⁰.
4. Voltage across the resistor R₁ is proportional to the current flowing through R₁.
   $R_1 = 270\ \Omega$
   
   Hence by Ohm's law, voltage across R₁, V_R is given by
   
   $V_R = I_TR_1$
   $V_R = 3.19\ \angle30.52^0\times 270\ mV$
   $V_R = 861.3\ \angle30.52^0\ mV$
   
   Therefore voltage across R₁ is
   $V_R = 861.3\ \angle30.52^0\ mV$
   
5. From the above values, the phases of I_T and V_R are given by
   $\angle V_R = 30.52^0$
   $\angle I_T = 30.52^0$
   
   Hence phase difference is zero since both of them are in phase.
   
   Hence we verified that the voltage across R₁ is in phase with the current flowing through it.