# Q = Integral Of Idt,Where Q Is The, Export Administration Regulations

Derivatives and integrals are widely used to describe transient processes in electric circuits. Below, we look at some typical problems that can be solved using integration. We confine ourselves to consideration of first order circuits.

Đang xem: Q = integral of idt

### Relationship Between Charge and Current

Electric current (I) is defined as the rate of flow of charge (Q) and is expressed by the derivative

Turning the equation the other way round, we get the integral formula

which represents the amount of charge passing through the wire between the times (t = {t_1}) and (t = {t_2}.)

### RC Circuit

A simple series RC Circuit is an electric circuit composed of a resistor and a capacitor.

Figure 1.

After the switch is closed at time (t = 0,) the current begins to flow across the circuit. The voltage across the resistor is given by the Ohm’s law:

<{V_R}left( t ight) = Ileft( t ight)R.>

The voltage across the capacitor is expressed by the integral

<{V_C}left( t ight) = frac{1}{C}intlimits_0^t {Ileft( s ight)ds} ,>

where (C) is a capacitance value, (s) is the internal variable of integration.

By the Kirchhoff’s voltage law (KVL), we can write

<{V_R}left( t ight) + {V_C}left( t ight) = varepsilon ,>

where (varepsilon) is the electromotive force (emf) of the power supply (we assume that (varepsilon) is constant).

Hence,

By differentiating with respect to (t,) we can convert this integral equation into a linear differential equation:

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which has the solution in the form

Figure 2.

The time constant ( au = RC) here determines how quickly the transient process in the circuit occurs.

### RL Circuit

A simple RL Circuit has a resistor and an inductor connected in series.

Figure 3.

When the switch at time (t = 0) is closed, a constant emf (varepsilon) is applied and the current (I) begins to flow across the circuit.

Similarly to the previous section, the voltage across the resistor is given by

<{V_R}left( t ight) = Ileft( t ight)R.>

The voltage across the inductor is expressed by the derivative

<{V_L}left( t ight) = Lfrac{{dI}}{{dt}}.>

So, by KVL,

<{V_R}left( t ight) + {V_L}left( t ight) = varepsilon ,>

or

Integrating this linear differential equation with the initial condition (Ileft( {t = 0}
ight) = 0) gives the following solution:

Figure 4.

We see that the time constant for an RL circuit is given by ( au = large{frac{L}{R}}
ormalsize.)

### Power and Energy

Electric energy (E,) measured in joules (J), is a form of energy that results from kinetic or potential energy possessed by electric charges.

Electric power (P,) measured in watts (W), is the rate at which electric energy is transferred by an electric circuit.

The power dissipated in a direct current (left({DC}
ight)) circuit element is given by the formula

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where (V) is the voltage across the element and (I) is the current in the circuit.

In particular, if the power is dissipated in a resistor of resistance (R,) then

The energy dissipated by a (DC) circuit element during the time period (left< {0,t} ight>) is given by

When the voltage and current change in time, the instantaneous power is defined as

In this case, the energy dissipated during the time interval (left< {0,t} ight>) is given by the integral

where (s) is the internal variable of integration.

### Energy Stored in a Capacitor

Moving a small charge (dq) from one plate of a capacitor to the other requires the work

where (C) is the capacitance and (q) is the current charge of the capacitor.

Integrating from (q = 0) to (q = Q) gives the total energy stored in the capacitor:

<{{E_C} = intlimits_0^Q {dW} }={ intlimits_0^Q {frac{q}{C}dq} }={ frac{{{Q^2}}}{{2C}} }={ frac{{C{V^2}}}{2}.}>

Figure 5.

### Energy Stored in an Inductor

Increasing the current in an inductor by a small value of (di) requires the work

<{dW = Pdt }={ – varepsilon idt }={ iLfrac{{di}}{{dt}}dt }={ Lidi.}>

Integrating from (i = 0) to (i = I) gives the total energy stored in the inductor:

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