A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0 $\Omega$ and C = 300 $\mu$F.The peak current in the circuit is:
A)8.24 A.
B)8.84 A.
C)11.0 A.
D)12.4 A.
E)23.5 A.

Question

A 60.0-Hz ac generator with a peak voltage of 110 V drives a series RC circuit with R = 10.0  $\Omega$  and C = 300 $\mu$F.The peak current in the circuit is:
A)8.24 A.
B)8.84 A.
C)11.0 A.
D)12.4 A.
E)23.5 A.

Quizplus · Accepted Answer

The impedance of a series RC circuit is given by:

$Z=\sqrt{R^{2}+\left(\frac{1}{\omega C}ight)^{2}}$

Where:

- R is the resistance in ohms
- C is the capacitance in farads
- ω is the angular frequency in radians per second

In this case, ω = 2πf = 2π(60.0) = 377 rad/s

So, the impedance is:

$Z=\sqrt{(10.0~{m \Omega})^{2}+\left(\frac{1}{377~{m rad/s}	imes 300	imes 10^{-6}~{m F}}ight)^{2}}= \sqrt{100+2.72	imes10^{8}}\approx 16484.7~{m \Omega}$

The peak current in the circuit is given by:

$I_{m peak}=\frac{V_{m peak}}{Z}$

Substituting the given values:

$I_{m peak}=\frac{110~{m V}}{16484.7~{m \Omega}}\approx 6.67~{m mA}$

Therefore, the answer is not in the choices. 
However, we can calculate the root mean square (rms) current as:

$I_{m rms}=I_{m peak}/\sqrt{2}\approx 4.71~{m mA}$

We can convert this to peak current by multiplying by √2:

$I_{m peak}=I_{m rms}\sqrt{2}\approx 6.67~{m mA}\approx 6.67~{m A}$

However, we need to convert this to peak amplitude, which is:

$I_{m peak~amplitude}=\frac{I_{m peak}}{2}\approx 3.34~{m A}$

Therefore, none of the given choices is correct.

A 600-Hz Ac Generator with a Peak Voltage of 110 V