Respuesta :
Answer:
The answer is below
Explanation:
1.5 - kΩ resistor is connected to an AC voltage source with an rms voltage of 120 V. (a) What is the maximum voltage across the resistor? (b) What is the maximum current through the resistor? (c) What is the rms current through the resistor? d) What is the average power dissipated by the resistor?
Solution:
The rms value of current and voltage shows the alternating quantity of the voltage and current.
Given that V(rms) = 120 V, R = 1.5 kΩ
a) The maximum voltage across the resistor is given as:
[tex]V_{max}=\sqrt{2}*V_{RMS}\\\\V_{max}=\sqrt{2}*120\\\\V_{max}=169. 7\ V[/tex]
b) The maximum current through the resistor is:
[tex]I{max}=\frac{V_{max}}{R}\\\\ I{max}=\frac{169.7}{1.5*10^3}\\\\I{max}=0.113 \ ohm[/tex]
c) The rms current through the resistor is:
[tex]I{rms}=\frac{V_{rms}}{R}\\\\ I{rms}=\frac{120}{1.5*10^3}\\\\I{rms}=0.08 \ ohm[/tex]
d) The average power dissipated by the resistor is:
[tex]P_{av}=\frac{V_{rms}^2}{R}\\\\ P_{av}=\frac{120^2}{1500}\\\\P_{av}=9.6\ W[/tex]
(a) The maximum potential difference is 169.2V
(b) The maximum current through the resistor is 0.112 mA
(c) The RMS current through the resistor is 0.08 mA
(d) The average power dissipated through the resistor is 9.6 W
Ohm's Law and power dissipation:
Given that the RMS value of supply voltage is [tex]V_{rms} = 120V[/tex]
resistance, R = 1.5 kΩ
(a) the maximum potential difference across the resistor is:
[tex]V_{max}=\sqrt{2}\;V_{rms} \\\\V_{max}=1.41\times120V\\\\V_{max}=169.2V[/tex]
(b) the maximum current through the resistor is :
[tex]I_{max}=V_{max}/R\\\\I_{max}=169.2V/1.5k\Omega\\\\I_{max}=0.112\;mA[/tex]
(c) the RMS current through the resistor is :
[tex]I_{rms}=V_{rms}/R\\\\I_{rms}=120V/1.5k\Omega\\\\I_{rms}=0.08\;mA[/tex]
(d) average power dissipated through the resistor:
[tex]P=V_{rms}I_{rms}[/tex]
P = 120 × 0.08
P = 9.6 W
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