Quiz 2 Hints

2. P_AV = ½·i_m^{2 ·}R/2= ½·Vm²/R = 75.0 W
Find R for both parts.

9.
(a) Use equations 33.9 and 33.10.
(b) Use equations 33.9 and 33.10. Find w given the maximum current.
Note:

18. q = CV so that i = dq/dt = C·dV/dt. Now, V = V_m·sin wt.
Thus, i = wC·V_m·cos wt. Note: q = 0 when t = 0 as stated in the problem, since the energy is zero at t = 0. Recall that U_C = q²/2C.
Note:

Find i at the time t given in the problem.
Note:

24. The problem represents a circuit with R and C in series without an inductor in the circuit! R = 50 K-ohms,
C = 20.0 pF .

Note that Z is like equation 33.23 without the presence of an inductor L.
Remember to divide V_mR by the square root of 2 to get the RMS value.
You might want to skim section 32.4 on transformers to see why the voltage is stepped up from 120 V to 5000 V.

26. Hint: Refer to fig 33.11. The diagram will look something like that, since the phase angle that you calculate will be positive for this inductive circuit.
Calculate the following:

Then draw a diagram like fig. 33.11 using the given value of R . The Z phasor lies in the first quadrant. Comment: For practice, draw the diagram corresponding to the circuit parameters of ex. 33.5, a capacitive circuit. Note that in this case, the phase angle is negative, which means the Z phasor will lie in the fourth quadrant.

28.
The analysis is basically the same as if we had a sine function, where if

then,

Note that in this problem we are given RMS values defined in the notation of the book:

(a)

Find i_m , the maximum current. Use eqn. 33.23 to find Z.

(b)

Note:

32. (a) Write the expression for Z without the presence of
of a capacitor:

(b)

Find the angle by using the equations I wrote in lecture, Monday Feb. 5. Note that you can use these same equations with the value of X_C = 0. There is no capacitor !!
(c)

This the condition for resonance!! That means X_L = X_C. Solve for the value of capacitance C such that X_L = X_C at the given frequency.
Note:

(d) In general, P_AV =i²_RMSR where i_RMS = V_RMS/Z
Before the addition of the capacitor:

and V_RMS = 120 V
Calculate P_AVbefore the addition of the capacitor using the above values of C and L . Call this P_AV1

After the addition of the capacitor, Z = R and the average power is P_{AV2 .} Note: P_AV2= i²_RMSR where i_RMS = V_RMS/R where V_RMS has a new value after the addition of the capacitor.

Write:
P_AV1 = P_AV2 or

P_AV1 = i²_RMSR or

P_AV1 =(V_RMS/R)²·R or

P_AV1 = V²_RMS/R. Solve for V_RMS.

39. The instantaneous rate of heat dissipated in the circuit is the power delivered to the resistor:

The average value of (sine)²is one-half.
Thus, the average power in the resistor is given by:

The, heat dissipated in one period is given by the product of the average power and the period:

In this problem, the operating angular frequency is given as twice the resonant frequency:

So, all you need to do to complete the problem is to find i_mwhen

Here is how you do this. Remember that:

To finish the problem all you have to do is evaluate Z when

Then substitute into the above expression for the maximum current. Finally, substitute into the above expression for the heat in one period. Note: The book gives the rms value for the voltage source:

.
Be careful !