Kirchhoff\'s law states that the algebraic sum of the electromotive forces aroun
ID: 2942471 • Letter: K
Question
Kirchhoff's law states that the algebraic sum of the electromotive forces around a closed circuit is zero.-A source of electromotive force (emf), E, (such as a battery) drives electric charge and produces a current,I.
-A resistor of resistance R which opposes current by producing a drop in emf of magnitude (Ohm's Law).
-An inductor of inductance L which opposes any change in current by producing a drop in emf of magnitude
Suppose you have a circuit with a battery, an inductor, and a resistor.
A)Write the initial value problem for this situation where and emf is a function of
time E(t).
B)Use the differential equation to explain why Ohm's law is satised whenever the current is at a maximum
or minimum.
C)Use the differential equation to show that the emf is increasing when the current is at a minimum and
decreasing when it is at a maximum. Hint: Differentiate your differential equation with respect
to time remembering that I,E are both functions of time
Explanation / Answer
Kirchhoff's law is total voltage = 0 V(battery) + V(resistor) + V (inductor) = 0 E(t) - IR - L dI/dt = 0 (V = IR for resister) Inductor = inductance * change in current over change in time) A) so then E(t) = IR + L dI/dt where I is current, r is resistance, t is time, l is inductance B) When current is max or minimum dI/dt = 0 because it is an inflection point This makes the equation (E(t) = IR + L dI/dt) = E(t) = IR + L(0) = E(t) = IR because E(t) = V of the battery, V = IR C) second differentiation = dE/dT = R*dI/dt + L d^2 I/dt^2 At a maximum, the system is concave down dI/dt = 0, d^2I/dt^2 = negative dE/dT = R*(0) + L (negative) = L (negitive value) so at a maximum, dE/dT is negative, so emf is decreasing because it's derivative is negative for minimum the system is concave up dI/dt = 0, d^2I/dt^2 = positive dE/dT = R*(0) + L (positive) = L (positive value) so emf is increasing because its derivative is positive so at a minimum, dE/dT is positive, so emf is increasing
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.