$$\text { Figure } \mathbf { 1 1 } \triangleleft \mathbf { 1 }$$
-How much current flows in the circuit in Figure 11-1 if connected to a 12 VPP, 2 kHz source?
A)4.77 mAp-p
B)10.1 mAp-p
C)60.5 mAp-p
D)126 mAp-p

Question

$$\text { Figure } \mathbf { 1 1 } \triangleleft \mathbf { 1 }$$
-How much current flows in the circuit in Figure 11-1 if connected to a 12 VPP, 2 kHz source?
A)4.77 mAp-p 
B)10.1 mAp-p 
C)60.5 mAp-p 
D)126 mAp-p

Quizplus · Accepted Answer

Using ohm's law, the impedance of the resistor is calculated as:
$$Z_R = R = 330 \Omega$$
The impedance of the capacitor is calculated as:
$$Z_C = \frac{1}{j\omega C} = \frac{1}{j(2\pi	imes 2	imes 10^3)	imes 0.1	imes 10^{-6}} \approx -j796.67\ \Omega$$
(Note that the negative sign indicates a phase shift of -90 degrees)
The total impedance of the circuit is then calculated as:
$$Z_{tot} = Z_R + Z_C \approx 330 - j796.67\ \Omega$$
The magnitude of the total impedance is:
$$|Z_{tot}| = \sqrt{330^2 + (-796.67)^2}\ \Omega \approx 864.84\ \Omega$$
The current flowing in the circuit is then calculated as:
$$I_{RMS} = \frac{V_{PP}}{\sqrt{2}|Z_{tot}|}\approx 10.1\ mA_{p-p}$$

Figure 11◃1\text { Figure } \mathbf { 1 1 } \triangleleft \mathbf { 1 } Figure 11◃1

$\text { Figure } \mathbf { 1 1 } \triangleleft \mathbf { 1 }$