Answer:
a) 1857.6 W
b) 3713.036 W
Explanation:
Given:
[tex]V_{rms}=240V\\R=31\Omega[/tex]
Average current, [tex]I=\frac{240V}{31\Omega}=7.74 A[/tex]
Average power, [tex]P=VI=240\times 7.74=1857.6W[/tex]
Peak current:
[tex]I_p=\sqrt2 I\\I_p=\sqrt 2 \times 7.74=10.94A\\[/tex]
Peak Voltage:
[tex]V_p=\sqrt2 V\\V_p=\sqrt2 \times 240 =339.4 V[/tex]
Maximum value of instantaneous Power is
[tex]P_{max}=I_p\times V_p\\P_{max}=10.94\times 339.4 = 3713.036 W[/tex]
The average power be "1857.6 W" and the max value of the instantaneous power be "3713.036 W".
Given values are:
Resistance, [tex]R = 31 \ \Omega[/tex]
Voltage, [tex]V_{rms} = 240 \ V[/tex]
The average current be:
→ [tex]I = \frac{240}{31}[/tex]
[tex]= 7.74 \ A[/tex]
PART-A:
The average power be:
→ [tex]P= VI[/tex]
[tex]= 240\times 7.74[/tex]
[tex]= 1857.6 \ W[/tex]
Now,
The peak current be:
→ [tex]I_p = \sqrt{2 }I[/tex]
[tex]= \sqrt{2}\times 7.74[/tex]
[tex]= 10.94 \ A[/tex]
The peak voltage be:
→ [tex]V_p = \sqrt{2} V[/tex]
[tex]= \sqrt{2}\times 240[/tex]
[tex]= 339.4 \ V[/tex]
PART-B:
The max. value of instantaneous power be:
→ [tex]P_{max} = I_p\times V_p[/tex]
[tex]= 10.94\times 339.4[/tex]
[tex]= 3713.036 \ W[/tex]
Thus the answer above is right.
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