# Root mean square value

General formula for RMS – root mean square

$$F_{RMS} = \sqrt{\frac{1}{T}\int_{0}^{T}{f^{2}(t)\cdot dt}}$$

The RMS value interpretation based on the electrical current root mean square value.

$$W = \int{p \cdot dt}$$
$$W = \int{u \cdot i \cdot dt}$$
$$W = R\cdot \int{ i^{2} \cdot dt}$$

where:
W – work
p – temporal electric power
$$p = u \cdot i$$
u – temporal electric voltage
i – temporal electric current

If the electrical current is periodical then

$$W_{T} = R\cdot \int_{0}^{T}{ i^{2} \cdot dt}$$

Root mean square value of the alternate electric current is an equivalent direct electric current which will produce exactly same amount of heat.

$$R\cdot \int_{0}^{T}{ i^{2} \cdot dt} = R \cdot I^{2} \cdot T$$
Dividing both sides of the above equation by resistance R
$$\int_{0}^{T}{ i^{2} \cdot dt} = I^{2} \cdot T$$
Next swapping sides in order to find relation for the DC electrical current I
$$I^{2} \cdot T = \int_{0}^{T}{ i^{2} \cdot dt}$$
Finally following is received

Root mean square value for the alternate electric current
$$I_{RMS} = \sqrt{\frac{1}{T}\cdot \int_{0}^{T}{ i^{2} \cdot dt}}$$

Root mean square value for the alternate electric voltage
$$U_{RMS} = \sqrt{\frac{1}{T}\cdot \int_{0}^{T}{ u^{2} \cdot dt}}$$

# A maximum power on a load in an electric DC circuit

A series electric DC circuit is being considered to do math. The circuit contains a voltage source with its internal resistance and a load resistors. Resistance of wires is omitted in this example.

The voltage equation for the circuit above is following
$$E - R_i \cdot I - R_o \cdot I = 0$$

Thereby the current is given by the equation
$$I = \frac{E}{ R_i + R_o }$$

A general formula for power is
$$P = U \cdot I$$

Knowing that the voltage on the resistor Ro is given by the Ohm’s law relation
$$U(R_o)= I \cdot R_o$$

A formula for power on the resistor Ro is following
$$P(R_o)= I^2 \cdot R_o$$

After plugging in the relation for current
$$P(R_o)= (\frac{E}{ R_i + R_o })^2 \cdot R_o$$

The extremum of function P(Ro) needs to be found in order to find which magnitude of the resistor Ro will be give the maximum power.

The first derivative of equation of power on the resistor Ro is following
$$\frac{dP(R_o)}{dR_o}= \frac{d}{dR_o}((\frac{E}{ R_i + R_o })^2 \cdot R_o)$$

After doing the math, the theorems for derivatives of the multiplication of functions and internal and external functions were applied
$$\frac{dP(R_o)}{dR_o}= -2\cdot \frac{E}{ R_i + R_o } \cdot \frac{E}{ (R_i + R_o)^2 } \cdot R_o + (\frac{E}{ R_i + R_o })^2$$

In order to find the extremum the point where the derivative is equal to 0 needs to be calculated
$$0= -2\cdot \frac{E}{ R_i + R_o } \cdot \frac{E}{ (R_i + R_o)^2 } \cdot R_o + (\frac{E}{ R_i + R_o })^2$$

After rearranging the elements in equation above and doing some math
$$R_o = R_i$$

Finally, the maximum power on the load resistor Ro is given by formula
$$P(R_o)_{max} = \frac{E^2}{4 \cdot R_i}$$

# A volume of an iceberg’s tip

From the idiom “the tip of the iceberg” to the terrible disaster of the Titanic on the April 10th 1912 which was cause by the ship’s collision with the iceberg in the Atlantic Ocean. In this article a strict relation between the whole iceberg and its tip is calculated with application of the widely known Archimedes principle.

The Archimedes principle describes the force which works on a material body which is submerged in other fluid. In general a value of this buoyancy force is equal to equivalent of the gravity force which was pushed out by the submerged material body; a direction of the buoyancy force is opposite to the gravity force.

$$\vec{F_{b}} = -\vec{g} \cdot d \cdot V$$

The iceberg presented above is in the static equilibrium, therefore, the vector of gravity force G is compensated by the vector of buoyancy force . Assuming that a chosen coordinate system shows that positive direction is upwards then:

$$\vec{F_{b}} - \vec{G}= \vec{0}$$

Previous equation, in scalar representation, can be written as

$$g \cdot d_{H_{2}O-LIQUID} \cdot (V - V') = \cdot d_{H_{2}O-ICE (SOLID)} \cdot V$$

After taking into account the numerical values for water’s density in solid and liquid state of condensation. It can be shown with quite good accuracy

$$V' = 0.0833 \cdot V$$

Thereby the tip of the iceberg constitutes less than the 90 percent of the iceberg. More detailed calculations are available at the Archimedes principle – example 1.

# Analysis of an AC electrical circuit

A considered below AC electric circuit is composed of an AC voltage source, a resistor, a capacitor, an AC current source and an inductive coil. The nodal analysis method is being applied to calculate currents in the circuit’s branches. It is recalled that in the nodal analysis method it is always assumed that an electrical potential of one of nodes is equal to 0.

Since the considered electric circuit is an AC circuit following formulas are to be applied:

$$\underline{Z} = \frac{1}{\underline{Y}}$$ $$\underline{Y} = \frac{1}{\underline{Z}}$$ $$\underline{I}_1$$ $$\sum{(\underline{I}_s)_a} = \underline{Y}_{R1C1} \cdot \underline{V}_{s1} + \underline{I}_{s2} = \underline{V}_a \cdot ( \underline{Y}_{R1C1} + \underline{Y}_{L1} ) - \underline{V}_b \cdot ( \underline{Y}_{R1C1} + \underline{Y}_{L1} )$$ $$\underline{Y}_{R1C1} = \frac{1}{R1 - j \cdot \frac{1}{\omega \cdot C1}}$$ $$\underline{Y}_{L1} = \frac{1}{\omega \cdot L1}$$

The further calculations related to the present example are available – Node voltage method example 2.

# A free fall of a material point

A free fall of a material point is being examined. In order to find the solution of the considered example formulas for an accelerated motion and kinetic and potential energy in the gravitational field are used.

$$v = v_0 + a \cdot t$$
$$s = v_0 \cdot t + \frac{a \cdot t^2}{2}$$
$$E_k = \frac{m \cdot v^2}{2}$$
$$E_p = m \cdot g \cdot h$$

A free fall of a material point

# Transmittance and state space representation equations designation

The considered electrical circuit is analyzed as an object which is to be controlled automatically, therefore, its transmittance and state space representation equations will be written down.

Transmittance is given by formula

$$H(s) = \frac{Y(s)}{U(s)}$$

State space representation equations

$$\dot{ \textbf{x} } = A \cdot \textbf{x} + B \cdot \textbf{u}$$ $$\textbf{y} = C \cdot \textbf{x} + D \cdot \textbf{u}$$

An example which is solved step by step for the considered electric circuit can be found here

Transmittance and state space representation

# An electrical circuit solved with the branch current method

In the examined electric circuit the currents and voltages will be computed. The Kirchhoff’s current law (KCL) and the Kirchhoff’s voltage law (KVL) will be employed.

branch current method – electric circuit

# A dynamic model of a series RLC circuit

An example where a transfer function H(s) and the state space representation equations are defined for a series RLC circuit. As a result the dynamic model of the considered electrical system will be obtained in two forms. The considered electrical circuit is composed of three components: resitor, capacitor and inductor.

http://www.mbstudent.com/control-theory-state-space-representation-RLC-circuit-example-1.html

# Total capacitance

Capacitance C is a one of basic parameters of electric circuits next to resistance R and inductivity L. Capacitance C is defined as relation of charge Q to voltage V → C=Q/V. The measurement unit of capacitance is Farad → [C]=1F, Farad is a derived unit of SI system. Sometimes it is essential to calculate capacitance of electrical circuit which contains a few capacitors in its topology, therefore, it is often said that total capacitance of electrical circuit is computed. Sometimes during circuits analysis a subject is to calculate the total capacitance which is seen from specific circuit’s terminals.

Total capacitance