Home Work Solutions 11

Home Work Solutions 11

11-1 Using the loop rule, derive the differential equation for an LC circuit:

2 2

1 0

Ld q q

dt C 

Sol: The loop rule, for just two devices in the loop, reduces to the statement that the magnitude of the voltage across one of them must equal the magnitude of the voltage across the other. Consider that the capacitor has charge q and a voltage (which we’ll consider positive in this discussion) V = q/C. Consider at this moment that the current in the inductor at this moment is directed in such a way that the capacitor charge is increasing (so i = +dq/dt). Equation 30-35 then produces a positive result equal to the V across the capacitor: V = L(di/dt), and we interpret the fact that di/dt > 0 in this discussion to mean that d(dq/dt)/dt = d²q/dt² < 0 represents a “deceleration” of the charge-buildup process on the capacitor (since it is approaching its maximum value of charge). In this way we can “check” the signs in Eq. 31-11 (which states q/C =  L d²q/dt²) to make sure we have implemented the loop rule correctly.

11-2 A series circuit containing inductance L₁ and capacitance C₁ oscillates at angular frequency ω. A second series circuit, containing inductance L₂ and capacitance C₂, oscillates at the same angular frequency. In terms of ω, what is the angular frequency of oscillation of a series circuit containing all four of these elements? Neglect

resistance. (Hint: Use the formulas for equivalent capacitance and equivalent inductance; see Section 25-4 and Problem 47 in Chapter 30.)

Sol: For the first circuit  = (L1C1)^–1/2, and for the second one  = (L2C2)^–1/2. When the two circuits are connected in series, the new frequency is

       

   

eq eq 1 2 1 2 1 2 1 1 2 2 2 1 1 2

1 1 1 2 1 2

1 1 1

/ /

1 1

, /

L C L L C C C C L C C L C C C C

L C C C C C





   

   

 

 

where we use ^1  L C_{1 1}  L C ._{2 2}

11-3 An alternating source with a variable frequency, a capacitor with capacitance C, and a resistor with resistance R are connected in series. The following figure gives the impedance Z of the circuit versus the driving angular frequency ω_d; the curve reaches an asymptote of 500 Ω, and the horizontal scale is set by ω_ds = 300 rad/s. The figure also gives the reactance X_C for the capacitor versus ω_d. What are (a) R and (b) C?

Sol: (a) The circuit has a resistor and a capacitor (but no inductor). Since the capacitive reactance decreases with frequency, then the asymptotic value of Z must be the resistance: R = 500 .

(b) We describe three methods here (each using information from different points on the graph):

method 1: At d = 50 rad/s, we have Z  700 , which gives C = (d Z² - R² )^1

= 41 F.

method 2: At d = 50 rad/s, we have XC  500 , which gives C = (d XC)^1 = 40 F.

method 3: At d = 250 rad/s, we have XC  100 , which gives C = (d XC)^1

= 40 F.

11-4 An alternating source with a variable frequency, an inductor with inductance L, and a resistor with resistance R are connected in series. The following figure gives the impedance Z of the circuit versus the driving angular frequency ω_d, with the

horizontal axis scale set by ω_ds = 1600 rad/s. The figure also gives the reactance X_L for the inductor versus ω_d. What are (a) R and (b) L?

Sol: (a) Since Z = R² + XL

2 and XL = d L, then as d  0 we find Z  R = 40 .

(b) L = XL /d = slope = 60 mH.

11-5 A typical light dimmer used to dim the stage lights in a theater consists of a variable inductor L (whose inductance is adjustable between zero and Lmax) connected in series with a lightbulb B, as shown in the following figure. The electrical supply is 120 V (rms) at 60.0 Hz; the light-bulb is rated at 120 V, 1000 W. (a) What Lmax is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 1000 W? Assume that the resistance of the lightbulb is independent of its temperature. (b) Could one use a variable resistor (adjustable between zero and Rmax) instead of an inductor? (c) If so, what Rmax is required? (d) Why isn't this done?

Sol:

(a) The power consumed by the light bulb is P = I²R/2. So we must let P_max/P_min = (I/I_min)² = 4, or

I I_min









2















2

 Z_max Z_min









2

 R² 

 



R

















2

4.

We solve for L_max:

L

 3R

 ^

3 120 V  

^{/ 1000 W}

2  60.0 Hz  ^{ 6.62  10}

^H.

(b) Yes, one could use a variable resistor.

(c) Now we must let

R_max R_bulb R_bulb









2

4, or

R

   4  1 ^R

^{   1 } _{1000 W} ^{120 V }

^{ 14.4 .}

(d) This is not done because the resistors would consume, rather than temporarily store, electromagnetic energy.

11-6 The following figure shows an “autotransformer.” It consists of a single coil (with an iron core). Three taps Ti are provided. Between taps T1 and T2 there are 50 turns, and between taps T2 and T3 there are 800 turns. Any two taps can be chosen as the primary terminals, and any two taps can be chosen as the secondary terminals. For choices producing a step-up transformer, what are the (a) smallest, (b) second smallest, and (c) largest values of the ratio Vs/Vp? For a step-down transformer, what are the (d) smallest, (e) second smallest, and (f) largest values of Vs/Vp?

Sol:

For step-up transformer:

(a) The smallest value of the ratio V V_s/ _pis achieved by using T₂T₃ as primary and T₁T₃ as secondary coil: V₁₃/V₂₃ = (800 + 50)/800 = 1.07.

(b) The second smallest value of the ratio V V_s/ _pis achieved by using T₁T₂ as primary and T₂T₃ as secondary coil: V₂₃/V₁₃ = 800/200 = 4.00.

(c) The largest value of the ratio V V_s/ _pis achieved by using T₁T₂ as primary and T₁T₃ as secondary coil: V₁₃/V₁₂ = (800 + 50)/50 = 17.0.

For the step-down transformer, we simply exchange the primary and secondary coils in each of the three cases above.

(d) The smallest value of the ratio V V_s/ _pis 1/17.0 = 0.0588.

(e) The second smallest value of the ratio V V_s/ _pis 1/4.00 = 0.250.

(f) The largest value of the ratio V V_s/ _pis 1/1.07 = 0.935.