Question

4) Calculate load voltage, phase, load current, phase and load power for the tables below. (fill the tables for figure4 circuit) For table 1, use calculated maximum power CL and for table2 use calculated maximum power RL VoadPhaseVload lload (mA) RL PL (mW) 100Ω 3900 560Ω 8200 1000Ω 12000 1800Ω 3300Ω 1 0000Ω Table 1: Load Power with respect to RL (use maximum power CL for this table) LoadPower PL (mW) CL Vload (V)PhaseVload() Iload (mA) Phaselload() 1 nF 3.3 nF 10 nF 25 nF 33 nF 50 nF 100 nF Table 2: Load Power with respect to CL (use maximum power RL for this table) R1 1K LS 10mH load Vload CL Vin 20Vpp, sinusoidal,10KHz RL Figure 4

Answer 1

The condition for the maximum power is

$Z_L^*=Z_1$

$\Rightarrow (R_L+\frac{1}{j\omega C_L})^*=R_1+j\omega L_S$

$\Rightarrow R_L-\frac{1}{j\omega C_L}=R_1+j\omega L_S$

$\Rightarrow R_L+\frac{j}{\omega C_L}=R_1+j\omega L_S$

Comparing the real and the imaginary parts,

$R_L=R_1\ \&\ C_L=\frac{1}{\omega^2 L_S}$

From the given circuit, we have

$R_1=1K\Omega$

$\omega=2\pi f=2\pi\times10\times10^{3}=6.2832\times10^4rad/sec$

$L_S=10\times10^{-3}H$

So,

$R_L=1K\Omega$

$C_L=\frac{1}{(6.2832\times10^4)^2\times10\times10^{-3}}=25.33\times10^{-9}F=25.33nF$

From the given circuit, the load voltage can be obtained from the voltage division principle as

$V_{load}=V_{in}\frac{Z_L}{Z_L+Z_1}=V_{in}\frac{R_L+\frac{1}{j\omega C_L}}{R_1+j\omega L+R_L+\frac{1}{j\omega C_L}}$

The magnitude of the load voltage is

$\Rightarrow |V_{load}|=|V_{in}|\frac{\sqrt{R_L^2+\left(\frac{1}{\omega C_L} \right )^2}}{\sqrt{(R_1+R_L)^2+\left(\omega L-\frac{1}{\omega C_L} \right )^2}}$ ........Eq.1

The phase angle is

$\Rightarrow \angle V_{load}=\tan^{-1}\left(-\frac{1}{\omega C_LR_L} \right )-\tan^{-1}\left(\frac{\omega L_S-\frac{1}{\omega C_L}}{R_L+R_1} \right )$ .......Eq.2

The load current is given by

$I_{load}=V_{in}\frac{1}{Z_L+Z_1}=V_{in}\frac{1}{R_1+j\omega L+R_L+\frac{1}{j\omega C_L}}$

The magnitude of the load current is given by

$\Rightarrow |I_{load}|=|V_{in}|\frac{1}{\sqrt{(R_1+R_L)^2+\left(\omega L-\frac{1}{\omega C_L} \right )^2}}$ .......Eq.3

the phase angle of the load current is given by

$\Rightarrow \angle I_{load}=-\tan^{-1}\left(\frac{\omega L_S-\frac{1}{\omega C_L}}{R_L+R_1} \right )$ ........Eq.4

The load power is given by

$P_L=|I_{Load}|^2R_L=\frac{|V_{in}|^2R_L}{{(R_1+R_L)^2+\left(\omega L-\frac{1}{\omega C_L} \right )^2}}$ .......Eq.5

Note: Please note that the values of voltage and currents in the Eq.1 to Eq.5 are RMS values.

By using the Eq.1 to Eq.5 we can fill out the table 1 and table 2.

Table-1 $(C_L=25.33nF)$

RL	Vload(V)	PhaseVload(deg)	Iload(mA)	PhaseIload(deg)	Load Power(mW)
100	4.0899	-80.9567	6.4282	0	4.1322
390	3.7620	-58.1719	5.0871	0	10.0926
560	3.8150	-48.2905	4.5327	0	11.5056
820	4.0136	-37.4609	3.8852	0	12.3777
1000	4.1755	-32.1420	3.5355	0	12.5000
1200	4.3537	-27.6366	3.2141	0	12.3967
1800	4.8147	-19.2424	2.5254	0	11.4796
3300	5.5241	-10.7801	1.6444	0	8.9237
10000	6.4409	-3.593	0.6428	0	4.1322

Table-2 $(R_L=1K\Omega)$

CL(nF)	Vload(V)	PhaseVload(deg)	Iload(mA)	PhaseIload(deg)	Load Power(mW)
1	7.3139	-3.8583	0.4586	82.5464	0.2104
3.3	7.4948	-13.7782	1.5217	64.5078	2.3154
10	5.9873	-32.1419	3.1854	25.7162	10.1465
25	4.1812	-32.2438	3.5355	0.2378	12.4998
33	3.9148	-29.9235	3.5261	-4.1761	12.4337
50	3.6665	-26.4678	3.4938	-8.8110	12.2067
100	3.4854	-22.2449	3.4421	-13.2018	11.8480

Please note that the values of voltages and currents mentioned in the above tables are RMS values. If you want them to be peak to peak values multiply them with $2\sqrt{2}$