## Instantaneous Voltage and Current Values in a Series RLC Circuit

### Task number: 1543

An AC series circuit consists of a resistor with resistance of 90 Ω, a coil with inductance of 1.3 H and a capacitor with capacitance of 10 μF. The circuit is connected to an AC voltage source with amplitude of 100 V and frequency of 50 Hz. Write an equation for instantaneous values of voltage and current in the circuit, if the initial phase of the current is zero.

#### Given values

We write down the quantities and values given in the section Assignment

Amplitude of the alternating voltage in the circuit *U*_{m}= 100 VFrequency of the AC power source *f*= 50 HzResistance of the resistor *R*= 90 ΩInductance of the coil *L*= 1.3 HCapacitance of the capacitor *C*= 10 μFInitial current phase *φ*_{0i}= 0Quantities we want to determine:

Instantaneous current value *i*(*t*) = ? (A)Instantaneous voltage value *u*(*t*) = ? (V)#### Hint 1

The behaviour of alternating voltage and current can be described by a sine function.

Find the equations describing the instantaneous values of the alternating voltage and the current.

#### Hint 2

When evaluating these equations we need to know the current amplitude

*I*_{m}and a phase difference*φ*_{0u}between the voltage and the current. These we determine by using Ohm's law for alternating current.In the task Series RLC Circuit you can find a detailed explanation of the calculation of the current and the phase difference.

#### Determining amplitude of current

*I*_{m}We apply Ohm's law for a circuit with alternating current:

\[I_m=\frac{U_m}{Z}=\frac{U_m}{\sqrt{R^2+\left(X_L-X_C\right)^2}}=\frac{U_m}{\sqrt{R^2+\left(2 \pi f L - \frac{1}{2 \pi f C}\right)^2}} \]

We insert the given values:

\[I_m=\frac{100}{\sqrt{90^2+\left(2 \pi \cdot 50 {\cdot} 1{.}3 - \frac{1}{2 \pi \cdot 50 {\cdot} 10 \cdot 10^{-6}}\right)^2}} \, \mathrm{A} \] \[I_m \,\dot{=}\, 0{.}79 \, \mathrm{A}\]#### Determining phase difference between voltage and current

*φ*_{0u}We use the formula for phase difference expressed by impedance of each element.

\[ \mathrm{tg\,} \varphi_{0u} = \frac{X_L - X_C}{R}= \frac{\omega L- \frac{1}{ \omega C}}{R}= \frac{2 \pi f L- \frac{1}{ 2 \pi f C}}{R} \] \[\varphi_{0u} = \mathrm{arctg\,}\left(\frac{2 \pi f L- \frac{1}{ 2 \pi f C}}{R}\right)\]

Numerical solution:

\[\varphi_{0u} = \mathrm{arctg\,}\left(\frac{2 \pi f L- \frac{1}{ 2 \pi f C}}{R}\right)= \mathrm{arctg\,}\left(\frac{2\pi \cdot 50 {\cdot} 1.3 - \frac{1}{ 2 \pi \cdot 50 {\cdot} 10 \cdot 10^{-6}}}{90}\right)\] \[\varphi_{0u} \,\dot{=}\, 45 ^{\circ} = \frac{\pi}{4} \]#### Instantaneous values of alternating voltage and current

**Instantaneous value of alternating current:**The phase difference of the current is zero; the current amplitude

\[ i(t)=I_m \, \sin(\omega t + \varphi_{0i})=I_m\, \sin( 2 \pi f t + \varphi_{0i} ) \] \[ i(t)\,\dot{=}\,0.79 \cdot \sin(2 \pi 50 t)\,\mathrm A =0.79 \cdot \sin(100 \pi t)\,\mathrm A\]*I*_{m}has been calculated above.We know the voltage amplitude

\[ u(t)=U_m \, \sin(\omega t + \varphi_{0u})=U_m \, \sin( 2 \pi f t + \varphi_{0u} ) \] \[u(t)\,\dot{=}\,100 \cdot \sin( 2 \pi 50 t + \frac{\pi}{4} )\,\mathrm V = 100\cdot \sin( 100 \pi t + \frac{\pi}{4} )\,\mathrm V\]*U*_{m}and we have determined the phase difference.The time values inserted into the equations are converted into seconds.

#### Answer

The equations for instantaneous values of the voltage and the current in a series RLC circuit with AC have approximately the shape of:

\[ i(t)=0{.}79 \cdot \sin\left({100 \pi t}\right)\,\mathrm A\] \[u(t)=100 \cdot \sin\left( 100 \pi t + \frac{\pi}{4} \right)\,\mathrm V \]