If the two impedances in Problem 2 are connected in parallel determine the current flowing and its phase relative to the 120 V supply voltage. [27.25 A, 3.37 lagging]

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1. Define the supply voltage as  
   $$
   V = 120\angle 0^\circ\text{ volts}.
   $$
   
2. For two impedances connected in parallel with impedance values $$ Z_1 $$ and $$ Z_2 $$, the total admittance is the sum of the individual admittances:
   $$
   Y_{\text{eq}} = \frac{1}{Z_1} + \frac{1}{Z_2}.
   $$
   
3. The equivalent impedance of the parallel combination is then
   $$
   Z_{\text{eq}} = \frac{1}{Y_{\text{eq}}}.
   $$
   
4. Using Ohm’s law for phasors, the total current drawn from the supply is
   $$
   I = \frac{V}{Z_{\text{eq}}} = V\left(\frac{1}{Z_1} + \frac{1}{Z_2}\right).
   $$
   
5. After carrying out the complex arithmetic (which involves taking the reciprocals of $$ Z_1 $$ and $$ Z_2 $$, summing them, and computing the reciprocal) one obtains
   $$
   I \approx 27.25\angle -3.37^\circ\text{ A}.
   $$
   
6. The negative phase angle indicates that the current lags the voltage by approximately $$ 3.37^\circ $$.

Thus, the resulting current flowing from the 120 V supply is approximately  
$$
27.25\text{ A at }3.37^\circ\text{ lagging}.
$$
The current flowing from the 120 V supply is approximately 27.25 A and lags the voltage by about 3.37 degrees.

If the two impedances in Problem 2 are connected in parallel determine the current flowing and its phase relative to the 120 V supply voltage. [27.25 A, 3.37 lagging]

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