Q 3.5-5E The power generated or dissipate... [FREE SOLUTION]

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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 3: Q 3.5-5E (page 121)

The power generated or dissipated by a circuit element equals the voltage across the element times the current through the element. Show that the power dissipated by a resistor equall²R, the power associated with an inductorequals the derivative of $\frac{1}{2} {LI}^{2}$ and the power associated with a capacitor equals the derivative of $\frac{1}{2} C {E_{c}}^{2}$ .

Short Answer

Expert verified

The power dissipated by a circuit is $P_{R} = I^{2} (t) R$ .

The power dissipated by an inductor is $P_{L} = \frac{1}{2} \frac{d}{dt} [{LI}^{2} (t)]$

The power associated by a capacitor is $P_{C} = \frac{1}{2} \frac{d}{dt} [C {E_{C}}^{2} (t)]$

Step by step solution

Evaluate the power dissipated by a circuit

The power is given by the equation P = I(t)V(t).

Since the voltage across resister is

$V (t) = E_{R} (t) = RI (t) P_{R} = I (t) RI (t) P_{R} = I^{2} (t) R$

Hence, the power dissipated by a circuit is role="math" localid="1664225724130" $P_{R} = I^{2} (t) R$ .

Find the power dissipated by an inductor.

Since the voltage across the inductor is

$V (t) = E_{L} (t) = L \frac{dI (t)}{dt} P_{L} = I (t) L \frac{dI (t)}{dt} = \frac{L}{2} 2 I (t) \frac{dI (t)}{dt} P_{L} = \frac{d}{dt} [\frac{{LI}^{2} (t)}{2}] P_{L} = \frac{1}{2} \frac{d}{dt} [{LI}^{2} (t)]$

Hence, the power dissipated by an inductor is $P_{L} = \frac{1}{2} \frac{d}{dt} [{LI}^{2} (t)]$

Determine the power associated by a capacitor

Since the voltage across the capacitor is

$V (t) = E_{C} (t) = \frac{1}{C} q (t) q (t) = C E_{c} (t) I (t) = \frac{dq (t)}{dt} = \frac{d [{CE}_{c} (t)]}{dt} P_{C} = \frac{d [{CE}_{C} (t)]}{dt} \frac{q (t)}{C} = \frac{C}{2} 2 E_{C} (t) \frac{{dE}_{C} (t)}{dt} P_{C} = \frac{d}{dt} [\frac{C {E_{C}}^{2} (t)}{2}] P_{C} = \frac{1}{2} \frac{d}{dt} [C {E_{C}}^{2} (t)]$

Hence, the power associated by a capacitor is $P_{C} = \frac{1}{2} \frac{d}{dt} [C {E_{C}}^{2} (t)]$

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Short Answer

Step by step solution

Evaluate the power dissipated by a circuit

Find the power dissipated by an inductor.

Determine the power associated by a capacitor

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