.7
If you mean the side length of a square is half a meter, then you have 0.5 meters x 0.5 meters = 0.25 square meters.
I assume the question is about an exponent whose value is a half (or a multiple of a half). An exponent of a half is simply the square root. Similarly, an exponent of (1/3) is a cubed-root and so on.
RMS is a type of average. It is the "root of the mean of the squares". That is, the individual values are squared, the average is taken, and the square root of this is calculated. Since the "individual values" are often continuous - a typical example is a voltage, which continuously changes for example as a sine wave - integration must be used.
It could mean a cube tat has been cut into a half so that it is a cuboid. Or, it could mean a cube, each of whose sides is half a unit.
If you mean, the square root of 4 multiplied by the square root of 5, that is 2x5=10 If you mean, the square root of ( 4x25), that is the square root of 100 which is 10
RMS means root mean square of a sinusoidal wave form and the number that describe it is .741 of the peak average is ,639 of the peak
Generally, the rectified voltage (DC Voltage) is less than the supply voltage (230 VAC or 110 VAC). Therefore, there is a need to step down the mains voltage to the required value before rectification. AC Voltage is denoted by the Root Mean Square (RMS) value which is equal to the peak voltage of the sine wave divided by 1.4 (square root of 2). Therefore, the out put of a full wave rectifier with a smoothening filter (say, a condenser) will be about 1.4 times the RMS value of AC Voltage. For eg. if we need 12 VDC output from the rectifier, the AC Voltage output of the transformer (which will be the input to the rectifier) should be 12/1.4 ie. 8.6 V. However, since a stabilized DC power supply will usually have some kind of a voltage stabilizer, the output of the rectifier can be higher. Therefore, the transformer output can be 12 volts RMS in this case. The DC output will be roughly 12 x 1.4 = 16.8 V and the stabilized DC voltage can be maintained at 12 V DC irrespective of small fluctuations in the AC mains voltage.
The input voltage to a rectifier is less than the output voltage because the rectification process involves converting alternating current (AC) to direct current (DC), which typically has a peak voltage that can exceed the RMS (root mean square) value of the AC input. Additionally, during rectification, voltage drops occur across the diodes or other components, which can further reduce the effective input voltage. As a result, while the output voltage can be higher momentarily, the overall conversion and losses lead to a lower input voltage compared to the peak output voltage in a rectified waveform.
For an alternating voltage, the simple mean over a cycle would be zero. 'RMS' means 'root mean square', and is defined as the square root of the mean value of the square of the voltage, taken over a cycle. Thus whether the voltage is + or - , as it is in alternate half cycles, the value of its square is always positive, giving a real number for the square root. In fact the RMS value of voltage produces an RMS current which dissipates power at the same rate as a DC current of the same value. To find the RMS value of a sine wave with no DC offset, divide the peak value of the sine wave by square root of 2. **************************************************** Since the r.m.s. value of a sine wave is 1.414Vpk, and the mean voltage of a sine wave is 1.57Vpk, then, starting with the r.m.s. value: Vmean = (Vr.m.s. x 1.414) ÷ 1.57
If you actually mean rectifier (rather than regulator), then you can determine if it is performing its base function of converting alternating current to direct current by using a voltmeter. If the rectifier is functioning, you should read a percentage (which depends upon whether it is a half-wave or full-wave rectifier) of the AC peak input value on the DC range of a voltmeter. Using an oscilloscope, you can clearly view the half-wave or full-wave unidirectional (positive or negative only) pulses produced at the output of the rectifier. If the rectifier is blown and is conducting in both directions you will see nothing on a DC voltmeter range (the average value of an AC waveform is zero), and on an oscilloscope you will see the full peak-to-peak AC input waveform at its output.
RMS stands for Root Mean Square. Power is calculated as V2/R where V is the voltage and R is the resistive component of a load, This is easy toi calculate for a DC voltage, but how to calculate it for a sinusoidal voltage? The answer is to take all the instantaneous voltages in the sine wave, square them, take the mean of the squares, then take the square root of the result. This is defined as the "heating effect voltage". For a sine wave, this is 0.707 of the peak voltage.
The root mean square (RMS) voltage is a measure of the effective voltage of an alternating current. It is calculated by taking the square root of the average of the squares of the voltage values over a given period of time. This value represents the equivalent direct current voltage that would produce the same amount of power in a resistive load.
30VAC is not equal and will never be equal to 30VDC. If you rectify 30v ac to DC you will get a voltage lower than 30v because there would be a voltage drop across the rectifier=diode,capacitor,resistor and transistors or thyristor, depending on the make of the rectifier.AnswerAs the questioner makes no mention using a rectifier, yes, 30 V a.c. is exactly equivalent to 30 V d.c. This is because a.c. voltages are root-mean-square values (unless otherwise stated) which, by definition, correspond to d.c. voltages.
For a 12V peak voltage (V_peak), the peak-to-peak value (V_pp) is 24V, as it is twice the peak voltage (V_pp = 2 * V_peak). The root mean square (RMS) value is approximately 8.49V, calculated as V_rms = V_peak / √2. The half-cycle average voltage is about 7.64V, calculated as V_avg = (V_peak / π).
A clipper is a device designed to prevent the output of a circuit from exceeding a predetermined voltage level without distorting the remaining part of the applied waveform.A clipping circuit consists of linear elements like resistors and non-linear elements like junction diodes or transistors but it does not contain energy-storage elements like capacitors. Clipping circuits are used to select for purposes of transmission, that part of a signal wave form which lies above or below a certain reference voltage level.In half wave rectification of a single-phase supply, either the positive or negative half of the AC wave is passed, while the other half is blocked. Because only one half of the input waveform reaches the output, mean voltage is lower. Half-wave rectification requires a single diode in a single-phase supply , or three in a three-phase supply. Rectifiers yield a unidirectional but pulsating direct current; half-wave rectifiers produce far more ripple than full-wave rectifiers, and much more filtering is needed to eliminate harmonics of the AC frequency from the output.Half-wave rectifierThe no-load output DC voltage of an ideal half wave rectifier for a sinusoidal input voltage is:Where:Vdc, Vav - the DC or average output voltage,Vpeak, the peak value of the phase input voltages,Vrms, the root-mean-square value of output voltage.
If you mean the side length of a square is half a meter, then you have 0.5 meters x 0.5 meters = 0.25 square meters.
The effective voltage of an electrical circuit is the measure of the average voltage over a complete cycle of alternating current. It is also known as the root mean square (RMS) voltage.