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Derive a power balance equation for the RLand RCcircuits. (See Problem 5.) Discuss the significance of the signs of the three power terms.

Short Answer

Expert verified

The power balance equation for RL and RC circuit are ddt(12LI2)+RI2=E (t) IandRI2+ddt  12CEC2=E (t) I respectively.

Step by step solution

01

Important formula.

The governing differential equation for RC circuit is dqdt+qCR=ER.

02

Determine the power balance for RL circuit

The power dissipated by a circuit is PR=I2(t) Rand the power dissipated by an inductor is PL=12ddt  [LI2(t)] and the power associated by a capacitor is PC=12ddt[CEC2(t)].

So, the equation is

LdIdt+RI=E(t)LIdIdt+RI2=IE(t)ddt(12LI2)+RI2=E(t)I

Therefore, the power balance equation for RL circuit is ddt12LI2+RI2=E (t) I.

03

evaluate the power balance for RC circuit.

Rdqdt+qc=EI=dqdtRI+qC=E(t)RI2+qC=E(t)I

Further, solve the above expression

q=CECRI2+CECC.dqdt=E(t)IRI2+ddt(12CEC2)=E(t)I

Therefore, the power balance equation for RC circuit is RI2+ddt12CEC2=E (t) I

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Most popular questions from this chapter

Use the improved Euler’s method subroutine with step size h = 0.2 to approximate the solution toat the points x = 0, 0.2, 0.4, …., 2.0. Use your answers to make a rough sketch of the solution on [0, 2].

Local versus Global Error. In deriving formula (4) for Euler’s method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With

\({\bf{g(t) = f(t,f(t))}}\), this approximation can be written as \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt}} \approx {\bf{hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \)where \({\bf{h = }}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ - }}{{\bf{x}}_{\bf{n}}}\) .

  1. Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectangle approximation has error\(\left( {\bf{O}} \right){{\bf{h}}^{\bf{2}}}\); that is, for some constant M, \(\left| {\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} } \right| \le {\bf{M}}{{\bf{h}}^{\bf{2}}}\).This is called the local truncation error of the scheme. [Hint: Write \(\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {{\bf{g(t)dt - hg(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} {\bf{ = }}\int\limits_{{{\bf{x}}_{\bf{n}}}}^{{{\bf{x}}_{{\bf{n + 1}}}}} {\left[ {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right]{\bf{dt}}} \). Next, using the mean value theorem, show that\(\left| {{\bf{g(t)dt - g(}}{{\bf{x}}_{\bf{n}}}{\bf{)}}} \right| \le {\bf{B}}\left| {{\bf{t - }}{{\bf{x}}_{\bf{n}}}} \right|\) . Then integrate to obtain the error bound\(\left( {\frac{{\bf{B}}}{{\bf{2}}}} \right){{\bf{h}}^{\bf{2}}}\).]
  2. In applying Euler’s method, local truncation errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after n steps is (O)h. This is the global error, which is the same as the[ss1] [m2] convergence rate of Euler’s method.


Building Temperature.In Section 3.3 we modeled the temperature inside a building by the initial value problem (13)\(\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{M}}\,{\bf{(t) - T}}\,{\bf{(t)}}} \right]{\bf{ + H}}\,{\bf{(t) + U}}\,{\bf{(t),}}\,\,{\bf{T}}\,{\bf{(}}{{\bf{t}}_{\bf{o}}}{\bf{) = }}{{\bf{T}}_{\bf{o}}}\) , where M is the temperature outside the building, T is the temperature inside the building, H is the additional heating rate, U is the furnace heating or air conditioner cooling rate, K is a positive constant, and \({{\bf{T}}_{\bf{o}}}\) is the initial temperature at time \({{\bf{t}}_{\bf{o}}}\) . In a typical model, \({{\bf{t}}_{\bf{o}}}{\bf{ = 0}}\) (midnight),\({{\bf{T}}_{\bf{o}}}{\bf{ = 6}}{{\bf{5}}^{\bf{o}}}\), \({\bf{H}}\left( {\bf{t}} \right){\bf{ = 0}}{\bf{.1}}\), \({\bf{U(t) = 1}}{\bf{.5}}\left[ {{\bf{70 - T(t)}}} \right]\)and \({\bf{M(t) = 75 - 20cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}\) . The constant K is usually between\(\frac{{\bf{1}}}{{\bf{4}}}{\bf{and}}\frac{{\bf{1}}}{{\bf{2}}}\), depending on such things as insulation. To study the effect of insulating this building, consider the typical building described above and use the improved Euler’s method subroutine with\({\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}}\) to approximate the solution to (13) on the interval \(0 \le {\bf{t}} \le 24\) (1 day) for \({\bf{k = 0}}{\bf{.2,}}\,{\bf{0}}{\bf{.4}}\), and 0.6.

The Taylor method of order 2 can be used to approximate the solution to the initial value problem\({\bf{y' = y,y(0) = 1}}\) , at x= 1. Show that the approximation \({{\bf{y}}_{\bf{n}}}\)obtained by using the Taylor method of order 2 with the step size \(\frac{{\bf{1}}}{{\bf{n}}}\) is given by the formula\({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{n}}}\). The solution to the initial value problem is\({\bf{y = }}{{\bf{e}}^{\bf{x}}}\), so \({{\bf{y}}_{\bf{n}}}\)is an approximation to the constant e.

If the resistance in the RLcircuit of Figure 3.13(a) is zero, show that the current I (t) is directly proportional to the integral of the applied voltage E(t). Similarly, show that if the resistance in the RCcircuit of Figure 3.13(b) is zero, the current is directly proportional to the derivative of the applied voltage.

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