When an AC source is connected across a 25.0 2 resistor, the output voltage is given by AV = (130 V)sin(60xt). Determine the following quantities. (a) peak voltage (b) rms voltage V. (c) rms current
When an AC source is connected across a 25.0 2 resistor, the output voltage is given by AV = (130 V)sin(60xt). Determine the following quantities. (a) peak voltage (b) rms voltage V. (c) rms current
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![**Problem: AC Circuit Analysis**
When an AC source is connected across a 25.0 Ω resistor, the output voltage is given by:
\[
\Delta V = (130 \, \text{V})\sin(60\pi t)
\]
Determine the following quantities:
(a) **Peak Voltage**
[Answer box] V
(b) **RMS Voltage**
[Answer box] V
(c) **RMS Current**
[Answer box] A
(d) **Peak Current**
[Answer box] A
(e) **Find the Current when \( t = 0.0045 \, \text{s} \).**
[Answer box] A
---
**Explanation:**
- The peak voltage of the AC source is given directly as 130 V.
- The root mean square (RMS) voltage is calculated using the formula \( V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \).
- The RMS current is derived from Ohm's law, using \( I_{\text{rms}} = \frac{V_{\text{rms}}}{R} \) where \( R = 25.0 \, \Omega \).
- The peak current is calculated as \( I_{\text{peak}} = \frac{V_{\text{peak}}}{R} \).
- To find the instantaneous current at \( t = 0.0045 \, \text{s} \), substitute \( t \) into the voltage equation and use Ohm's Law to find the current.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6d45b7c-770b-4dbe-b3cf-74fe1dfded8d%2F503345bf-c35d-4ee4-9478-896f3d198793%2Fui5f1ce_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: AC Circuit Analysis**
When an AC source is connected across a 25.0 Ω resistor, the output voltage is given by:
\[
\Delta V = (130 \, \text{V})\sin(60\pi t)
\]
Determine the following quantities:
(a) **Peak Voltage**
[Answer box] V
(b) **RMS Voltage**
[Answer box] V
(c) **RMS Current**
[Answer box] A
(d) **Peak Current**
[Answer box] A
(e) **Find the Current when \( t = 0.0045 \, \text{s} \).**
[Answer box] A
---
**Explanation:**
- The peak voltage of the AC source is given directly as 130 V.
- The root mean square (RMS) voltage is calculated using the formula \( V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \).
- The RMS current is derived from Ohm's law, using \( I_{\text{rms}} = \frac{V_{\text{rms}}}{R} \) where \( R = 25.0 \, \Omega \).
- The peak current is calculated as \( I_{\text{peak}} = \frac{V_{\text{peak}}}{R} \).
- To find the instantaneous current at \( t = 0.0045 \, \text{s} \), substitute \( t \) into the voltage equation and use Ohm's Law to find the current.
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