Calculating Voltage Across R and C in Series

In summary, a resistor and capacitor in series with an AC voltage source have a time-dependent voltage across the capacitor given by V_{C}(t) = V_{C_0} sin(\omega*t). When R=3 k Ohms, C=100 pF, V_{C0}=100 mV, and angular frequency {\omega}= 10^5 rad/s, the voltage across the resistor (V_{R}) can be calculated using the equation V_{R} = V_{C_{0}}{\cdot}R{\cdot}\frac{1}{C{\cdot}{\omega}}. Understanding the dimensionless nature of radians is important in this calculation.
  • #1
themonk
16
0

Homework Statement


A resistor with resistance R and a capacitor with capacitance C are connected in series to an AC voltage source. The time-dependent voltage across the capacitor is given by V[tex]_{C}[/tex](t) = V[tex]_{C_0}[/tex] sin([tex]\omega[/tex]*t). (there are no superscripts, but they show up as super for me, though I wanted them to be subscripts)

If R=3 k Ohms , C=100 pF, V[tex]_{C0}[/tex]=100 mV, and angular frequency [tex]{\omega}[/tex]= 10^5 rad/s, what is V[tex]_{R}[/tex]?

Homework Equations



[tex]\frac{V_{C_{0}}{\cdot}R}{\frac{1}{C{\cdot}{\omega}}}[/tex]

The Attempt at a Solution


So I plugged in the variables with 3 k Ohms being 3000 [tex]\frac{m^{2}*kg}{s*C^{2}}[/tex], 100 pF being 1E-9 [tex]\frac{s^{2}*C^{2}}{m^{2}*kg}[/tex], 100 mV being 1 [tex]\frac{m^{2}*kg}{C*s^{2}}[/tex] and 10[tex]^{5}[/tex].

When I plugged it all into the equation I got .03 V which is 30 mV*rad. So first there was the problem of the radians being there there is no information about changing it into meters or other useful information. Do I need to use a trig function perhaps?

The hint I was given was : Use the equation obtained in Part B (the equation above) to work out the answer. Be careful of powers of ten in your calculation.

Thank you
 
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  • #2
hi themonk! :smile:

(have an omega: ω and try using the X2 icon just above the Reply box :wink:)
themonk said:
When I plugged it all into the equation I got .03 V which is 30 mV*rad. So first there was the problem of the radians being there there is no information about changing it into meters or other useful information. Do I need to use a trig function perhaps?

mmm … you really like breaking down your https://www.physicsforums.com/library.php?do=view_item&itemid=101", don't you? :smile:

radian is dimensionless (metre metre-1) …

you can just multiply any unit by it, and the result is in the same units

for example, if an object moves with position x = rcosωt, y = rsinωt, its velocity is (-ωrsinωt, ωrcosωt), which looks as if it is in units of metre radians per second, but is in fact only metres per second :wink:
 
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  • #3
Thanks tiny-tim for your advice and help! In the end I had converted 100 pF to 1E-9 F rather than 1E-10 so my answer was off by one degree (30 mV vs 3 mV). And I thought the radians would "disappear" somehow, so thanks again.
 

What is the formula for calculating voltage across R and C in series?

The formula for calculating voltage across R and C in series is V = VR + VC, where VR is the voltage across the resistor and VC is the voltage across the capacitor.

How do you calculate the voltage across a resistor in a series circuit?

To calculate the voltage across a resistor in a series circuit, use Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance of the resistor.

How do you calculate the voltage across a capacitor in a series circuit?

To calculate the voltage across a capacitor in a series circuit, use the equation VC = Q/C, where VC is the voltage across the capacitor, Q is the charge stored on the capacitor, and C is the capacitance of the capacitor.

Can you calculate the voltage across a resistor and capacitor in series if the frequency changes?

Yes, the voltage across a resistor and capacitor in series can be calculated for any frequency. However, the values of the resistor and capacitor may affect the voltage differently at different frequencies.

How does the voltage across a resistor and capacitor in series change with time?

The voltage across a resistor in a series circuit remains constant with time, while the voltage across a capacitor decreases as the capacitor charges up. Eventually, the voltage across the capacitor will reach the same value as the voltage across the resistor.

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