Understanding rms values and peak values of voltage/current

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• Peak Value of Output Voltage and Voltage Gain of an Amplifier

  22417 Peak Value of Output Voltage and Voltage Gain of an Amplifier A power amplifier when driven by a current source of 5mA rms and a source resistance of 8k, delivers 12.5W signal power to an 8 load resistance.

• Arc Waveform & AC Power

  That is why we have peak to peak and RMS (root mean square) value of the voltage when we deal with ac waveforms. Peak voltage = sqrt(2) x RMS voltage (sqrt(2) = 1.414) Peak to peak is twice the above.

• Finding rms current through Capacitor and Inductor

  f0 = 1/{2*pi*Sqrt(LC)} (8) Putting in the values to (8) we get f0 = 1/{2 x 3.142 x Sqrt(10^-2 x 0.025 x 10^-6)} ~= 10^4 Hz = 10 kHz Using knowledge of the driving voltage of a sine wave oscillator operating at 10 kHz the rms current through

• Electromagnetism problems

  Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise. 6) A series RCL circuit contains only a capacitor (C = 9.46 F), an inductor (L = 5.71 mH), and a generator (peak voltage = 79.2 V, frequency

• RMS and Line-to-Line Voltages

  All the fundamental and harmonic components are effective values corresponding to the steady-state operating condition, which are not the transient values (for example, transient values could be the peak value of the sine wave voltage waveform).

• Basic electrical circuit engineering

  I = 36/360 = 0.1 A or 100 mA (b) Now we replace the battery with a sine wave generator with peak voltage of V(peak) = 36V. We need to find the current that flows so we will first need to find the RMS Voltage.

• Calculating capacitance and amplitude of a voltage

  Reference: Fundamentals of Physics by Resnick, Halliday and Walker. Also, if you want to convert rms values of voltages and currents into peak values of either, multiply by the square root of 2.

• Power supplies regulated & unregulated problems and solutions

  to peak voltage across secondary coil of the transformer is √2 the rms voltage so V_SEC (p-p)=√2.V_SEC (rms)=√2×23=32.53V The peak voltage across the secondary coil of the transformer is ½ of the peak to peak value so we find

• DC power supply design

  Q2 solution: (a) The RMS ripple voltage is the product of the ripple factor (RF) and the average DC voltage value Vdc. Since RF = 0.005 and Vdc = 28 V, the RMS ripple voltage is 0.14 V.

• AC-LCR Circuits: Inductance, Capacitance and Frequency

  The effective value of the voltage or current is called root mean square value of rms value. It appears that in the question the voltage of the source given is the rms value of the voltage.