00:01
So if we start off with part a, since the three elements are connected in parallel, at any given instant in time, they will have all three have the same voltage drop across them.
00:13
So that is the voltages across each element will be in phase with the source.
00:19
So the current in the resistor is in phase of the voltage source with magnitude given by oms law.
00:26
So i, r of t is equal to v .0 over r, sine of wt.
00:40
For part b, the current through the inductor will lag behind the voltage by pi over 2, with magnitude equal to the voltage source divided by the inductive reactants.
00:52
So i, l of t is equal to v .0 over xl.
01:00
Sign of w t minus pi over two.
01:07
For c, the current through the capacitor leads the voltage by pi over two with magnitude equal to the voltage source divided by the capacitive resistant or reactants.
01:19
So that would be i .c of t equals v .0 over xc sine sine of w t plus pi over or two.
01:36
For part d, the total current is the sum of the currents through each element, and we use a phaser diagram to add the currents, as was used in section 30 -8, to add the voltages with different phases...