A capacitor (C) is placed in series in a circuit with a power supply and a resistor (R).A sin wave voltage input is supplied and the voltage across the resistor (proportional to the current) and across both the capacitor and resistor (voltage) observed on an oscilloscope.The period (T) and the time difference (to) between the signals is measured and the phase difference φ is calculated.A student obtains values of:T = 166 µsφ = 43 oR = 499 ΩDetermine the value of the capacitance in the circuit, C. (Ignore the negative sign in the equation in the lab manual.) Give your answer in SI units to three significant figures.Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9
Question
A capacitor (C) is placed in series in a circuit with a power supply and a resistor (R).A sin wave voltage input is supplied and the voltage across the resistor (proportional to the current) and across both the capacitor and resistor (voltage) observed on an oscilloscope.The period (T) and the time difference (to) between the signals is measured and the phase difference φ is calculated.A student obtains values of:T = 166 µsφ = 43 oR = 499 ΩDetermine the value of the capacitance in the circuit, C. (Ignore the negative sign in the equation in the lab manual.) Give your answer in SI units to three significant figures.Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9
Solution
To find the capacitance (C) in the circuit, we can use the formula for the phase difference in a RC circuit:
φ = arctan(1/(2πfRC))
where f is the frequency, R is the resistance, and C is the capacitance.
First, we need to convert the phase difference from degrees to radians, since the arctan function uses radians.
φ = 43 degrees = 43 * π/180 = 0.75 radians
Next, we need to find the frequency (f). The frequency is the reciprocal of the period (T):
f = 1/T = 1/(166E-6) = 6024 Hz
Now we can rearrange the formula for φ to solve for C:
C = 1/(2πfR*tan(φ))
Substituting the given values:
C = 1/(2π * 6024 Hz * 499 Ω * tan(0.75 radians))
C = 1.77E-9 F
So, the capacitance in the circuit is approximately 1.77 nF.
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