Question:
The instantaneous voltage through a circuit of impedance 10 \(\Omega\) is \(V=40\sin(50\pi t)\). The effective value of current is
The instantaneous voltage through a circuit of impedance 10 \(\Omega\) is \(V=40\sin(50\pi t)\). The effective value of current is
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Read the peak voltage and divide the rms value by the impedance.
Updated On: Nov 13, 2026
- \(2\sqrt{2}\) A
- \(4\sqrt{2}\) A
- 2 A
- 4 A
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The peak voltage is \(V_0=40\ \text{V}\) and the rms value is \(V_{rms}=\dfrac{V_0}{\sqrt{2}}\).
Step 2: Effective current
\(I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{40}{\sqrt{2}\times10}\).
Step 3: Calculation
\(I_{rms}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\ \text{A}\).
Final Answer:
The effective current is \(2\sqrt{2}\) A. \[ \boxed{\text{(A) }2\sqrt{2}\text{ A}} \]
The peak voltage is \(V_0=40\ \text{V}\) and the rms value is \(V_{rms}=\dfrac{V_0}{\sqrt{2}}\).
Step 2: Effective current
\(I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{40}{\sqrt{2}\times10}\).
Step 3: Calculation
\(I_{rms}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\ \text{A}\).
Final Answer:
The effective current is \(2\sqrt{2}\) A. \[ \boxed{\text{(A) }2\sqrt{2}\text{ A}} \]
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