a series circuit contains:an AV power source(15v(peak to peak)/1.5 kHz/ 0 Deg)a 100 ohm resistor and a 10mH inductor.Calculate the peak to peak voltage across the resistor and inductor and determine the phase angle for each with reference to input voltage. will be quick to award best answer if it is good quality! :)
Your question relies upon on the circuit layout. If the two aspects are in parallel then the voltage will equivalent the means furnish by way of fact they're one and a similar. different connections would have a diverse consequence. An inductor shops ac means and releases it back into the circuit throughout the time of each and every ac cycle. while voltage is utilized to an inductor present day flows slowly before each and every thing and then ramps up greater effectively storing means. while the provision voltage is decreased, it releases that saved means performing very similar to yet another means furnish.
calculate Inductive reactance as X? 2πfL in ohms You can use the peak-peak voltage for all the calculations, realizing that the results are in peak-peak also. Now calculate the Impedance Z √(R? + X??) angle of impedance is θ arctan (X?/R) Now the current as I E/Z Now the voltage across the resistor as E IR Now the voltage across the inductor as E IX? note that they do not add up to the total voltage. .
replace the inductor by its equivalent impedancejwlj(2*pi*f)*lj(2*3.14*1500) ohms now find the current.
Xl 2pi fl (6.28) X (1500Hz) X (.01h) 94.2 Ohm Angle between circuit Voltage and current arc tan 94.2/100 43.29 degrees Zt (94.2 Ohms) / (sine 43.29 degrees) 137.4 Ohms it Vt/Zt 15V/137.4 Ohm .109 Amps (i of R) (i of L) all in phase with each other. V of R (i of R) X (R) (.109A) X (100 Ohm) 10.9 Volts V of L (i of L) X (X of L) (.109A) X (94.2 Ohm) 10.3 Volts Using the input Voltage as reference zero, the fact that inductor current lags inductor Voltage by ninety degrees and the fact that the resistor current and inductor current are both in phase with the input current the Voltage phase relationships are as follows: The 15V Input Voltage is at zero degrees The 10.9 Volts across the resistor is at minus 43.29 degrees The 10.3 Volts across the inductor is at plus 46.71 degrees. It may be worth noting that the (V of R) X (cos 43.29 degrees) + (V of L) X (cos 46.71 degrees) The source Voltage (15Vpp). This just shows that the sum of the Voltage drops around this series circuit is actually equal to zero even though the sum of the absolute values of (V of R) and (V of L) is greater than the source Voltage of 15 Volts. I think that is what these type of problems is trying to illustrate.
