HW9 solutions

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Apr 3, 2024

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HW 9 Homework Solutions Points: 1 1. An inductor is a circuit element designed to have a specific value of inductance (L) *a) True b) False Solution: Terminology discussed in lecture #19 -------------------------------------------------------------- Points: 3 2. Two inductors are in series. What is the change in potential across both inductors if the inductance of #1 is 0.2, the inductance of #2 is 0.33, and the rate of change of current through #1 is -36? The values are given in SI units and you should find the answer in SI units. a) 63.57 b) 32.52 c) 39.39 d) 95.53 *e) 19.08 Solution: - When two elements are in series, the current through both is the same so I1=I2=I - The equivalent inductance of the two in series is Leq=L1+L2 and the potential difference across it is deltaV=-Leq*dIdt, where dIdt (shorthand for the derivative dI/dt) is the rate of change of the current. Thus: deltaV = -(L1+L2) * dIdt deltaV = -(0.2+0.33) * -36 = 19.08 -------------------------------------------------------------- Points: 3 3. A certain volume of space has a uniform magnetic field and a uniform electric field. What is the total stored energy if the magnitude of the magnetic field is 7.600E-07, the magnetic constant is 1.260E-06, the magnitude of the electric field is 38, the Coulomb constant is 9.000E+09, and the volume is 0.5? The values are given in SI units and you should find the answer in SI units. a) 2.942E-07 b) 5.154E-07 c) 7.256E-07 *d) 1.178E-07 e) 1.572E-07 Solution: - The energy stored in the magnetic field is Um=B^2*vol/(2*mu0), where B is the magnitude of the magnetic field, vol is the volume of space, and mu0 is the magnetic constant (1.26e-6) - The energy stored in the electric field is Ue=E^2*vol/(2*4*pi*k), where E is the magnitude of the electric field,vol is the volume of space, and k is the Coulomb constant (9e9) - Therefore the total stored energy is Utot = Um + Ue, which since we can factor out the common vol term, is equal to: Utot = ( B^2 / (2 * mu0) + E^2 / (2 * 4 * pi * k) ) * vol Utot = ( 7.600E-07^2 / (2 * 1.260E-06) + 38^2 / (2 * 4 * pi * 9.000E+09) ) * 0.5 = 1.178E-07

-------------------------------------------------------------- Points: 1 4. If the surface of a wire loop lies in the XY plane and a magnetic field is pointing in the +z direction and its magnitude is decreasing the loop will produce a magnetic field pointing in the +z direction *a) True b) False Solution: This can be determined by Lenz’s law, as discussed in lecture #19. The induced magnetic field opposes the change in magnetic flux. Since magnetic flux in the +z direction is decreasing, the induced field will point in the +z direction to increase the magnetic flux in this direction. -------------------------------------------------------------- Points: 3 5. A battery is first connected across a resistor and a certain constant current is observed. Then the resistor and an inductor are put in series in a loop circuit with no battery and a decreasing current is observed. What must the current through the resistor when it is attached to the battery be if the time it takes the current through the inductor to decrease from its initial value to 37% of that is 0.9341, the inductance of the inductor is 5.2, and the voltage of the battery is 5.4? The values are given in SI units and you should find the answer in SI units. a) 0.76 b) 0.49 c) 0.85 *d) 0.97 e) 0.43 Solution: - The current through the resistor when connected to the battery is calculated by applying Ohm's Law: I=Vb/R, where Vb is the voltage of the battery and R is the resistance of the resistor. Thus the resistance R=Vb/I - The time it takes the current in an RL circuit with no battery to decrease from its initial value to 37% of that is simply t=L/R (as 37% is a special value for which this is true). Substituting in the R=Vb/I from above gives: t = L * I / Vb Solve for I = (Vb * t) / L I = (5.4 * 0.9341) / 5.2 = 0.97 -------------------------------------------------------------- Points: 3 6. A resistor and inductor are in series in loop circuit. Assume there is an initial current at time zero. What must the time be if the current is 2.733, the initial current is 3.1, the resistance of the resistor is 0.27, and the inductance of the inductor is 1.5? The values are given in SI units and you should find the answer in SI units. a) 0.74 *b) 0.70 c) 0.89 d) 0.36 e) 0.47 Solution: The current at a point in time t, if the resistance is R and inductance L, is: I = I0 * exp(-R * t / L)

Solve for t = (-L / R) * log(I / I0) t = (-1.5 / 0.27) * log(2.733 / 3.1) = 0.7 Note: “log” refers to the “natural log” function, also sometimes written “ln” -------------------------------------------------------------- Points: 1 7. In a circuit having a resistor and inductor in series in a loop if there is an initial current I0 then the current will decrease to 37% of that value in a time t=L/R *a) True b) False Solution: This is a special case of the general solution for decreasing current vs. time, as discussed in lecture #20 -------------------------------------------------------------- Points: 1 8. In a circuit having an inductor and capacitor in series in a loop, if there is initially a charge on the capacitor then the charge will oscillate in time *a) True b) False Solution: This the behavior of the “LC circuit” discussed in lecture #20 -------------------------------------------------------------- Points: 3 9. Assume the current starts at zero in a loop circuit with a battery, resistor, and inductor in series. What must the maximum current be if the current at a certain point in time is 0.02427, the resistance of the resistor is 0.61, the time is 0.44, and the inductance of the inductor is 64? The values are given in SI units and you should find the answer in SI units. a) 8.9 b) 9.4 c) 6.2 d) 7.1 *e) 5.8 Solution: The current at a point in time t, if the resistance is R and inductance L, is: Ip = Imax * (1-exp(-R * t / L)) Solve for Imax = -Ip / (exp(-(R * t) / L) - 1) Imax = -0.02427 / (exp(-(0.61 * 0.44) / 64) - 1) = 5.8 -------------------------------------------------------------- Points: 3 10. A circuit consists of a capacitor and inductor in series in a loop and oscillates with a certain frequency. By what number would the frequency of oscillation be multiplied if the inductance of the inductor is divided by 2 and the capacitance of the capacitor is increased by a factor of 10? Assume all other parameters remain unchanged. a) A number >=0.04 and <0.12 b) A number >=0.12 and <0.28 *c) A number >=0.28 and <0.48 d) A number >=0.48 and <0.98 e) None of the above

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Solution: - The frequency of oscillation is f=1/(2*pi*(L*C)^(1/2)), where L is the inductance and C the capacitance - For this analysis 1/(2*pi) is a constant whose value doesn't matter since it remains unchanged, and therefore we can ignore this constant (pretend it is equal to 1) and consider only f=1/(L*C)^(1/2) - Since the variables are directly or indirectly proportional the analysis can be simplified by considering the case where the initial values are L=1, C=1, and the initial f = 1 / (L * C)^(1 / 2) = 1 / (1 * 1)^(1 / 2) = 1 - After the specified changes we have final values L=0.5, C=10, and f = 1 / (L * C)^(1 / 2) = 1 / (0.5 * 10)^(1 / 2) = 0.447 - Comparing this with the initial f=1 we see that f is multiplied by 0.447 -------------------------------------------------------------- Points: 3 11. Suppose a circuit contains two inductors in parallel. What is the rate of change of the current through #1 if the inductance of #1 is 0.3, the inductance of #2 is 0.16, and the rate of change of current through #2 is -24? The values are given in SI units and you should find the answer in SI units. a) -87.2 b) -56.4 c) -15.2 d) -38.6 *e) -12.8 Solution: - When two elements are in parallel, the potential difference deltaV across both is the same - For #1 deltaV = -L1 * dI1dt, where L1 is the inductance of #1 and dI1dt (stands for the derivative dI1/dt) is the rate of change of the current thru #1 - For #2 deltaV = -L2 * dI2dt, where L2 is the inductance of #2 and dI2dt (stands for the derivative dI2/dt) is the rate of change of the current thru #2 - Equating these two deltaV equations gives -L1 * dI1dt = -L2 * dI2dt, thus: dI1dt = (L2 / L1) * dI2dt dI1dt = (0.16 / 0.3) * -24 = -12.8 -------------------------------------------------------------- Points: 1 12. Energy is stored in an inductor if current is flowing through it *a) True b) False Solution: This can be seen from the formula U=L*I^2/2, where I is current, as discussed in lecture #20 -------------------------------------------------------------- Points: 1 13. If current in a wire loop changes in time this will result in a changing magnetic flux through the loop that will affect the potential difference across the same wire. *a) True

b) False Solution: This is the phenomenon that leads us to define “inductance”, as discussed in lecture #19. -------------------------------------------------------------- Points: 3 14. A solenoid inductor has current going through it. What is the energy stored in this inductor if the magnetic constant is 1.260E-06, the number of loops in the solenoid is 19, the area of each loop is 0.5573, the current is 78, and the length of the solenoid is 0.3119? The values are given in SI units and you should find the answer in SI units. a) 4.259 *b) 2.472 c) 6.141 d) 5.498 e) 9.976 Solution: - The energy stored in an inductor is U=L*I^2/2, where L is the inductance and I the current - The inductance of the solenoid is L=mu0*N^2*A/l, where mu0 is the magnetic constant, N the number of loops, A the area of each loop, and l the length of the solenoid - Putting the two equations together gives: U = mu0 * N^2 * A * I^2 / (2 * l) U = 1.260E-06 * 19^2 * 0.5573 * 78^2 / (2 * 0.3119) = 2.472