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In an L-C -R series circuit connected to...

In an L-C -R series circuit connected to an AC source `V=V_(0)sin(100pi(t)+(pi)/(6))``V_(R)=40V, V_(L)=40 and V_(C )=10V`, resistance `R=4Omega`
peak value of current is

A

`10sqrt(2) A`

B

`15sqrt(2) A`

C

`20sqrt(2) A`

D

`25sqrt(2) A`

Text Solution

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The correct Answer is:
To find the peak value of current in the given L-C-R series circuit connected to an AC source, we can follow these steps: ### Step 1: Understand the given values We are given: - \( V_R = 40 \, V \) (Voltage across the resistor) - \( V_L = 40 \, V \) (Voltage across the inductor) - \( V_C = 10 \, V \) (Voltage across the capacitor) - \( R = 4 \, \Omega \) (Resistance) - The AC voltage source is given by \( V = V_0 \sin(100 \pi t + \frac{\pi}{6}) \). ### Step 2: Calculate the RMS current using the voltage across the resistor The voltage across the resistor \( V_R \) is related to the RMS current \( I_{RMS} \) by Ohm's law: \[ V_R = I_{RMS} \cdot R \] Substituting the known values: \[ 40 = I_{RMS} \cdot 4 \] To find \( I_{RMS} \): \[ I_{RMS} = \frac{40}{4} = 10 \, A \] ### Step 3: Calculate the peak current The peak current \( I_{peak} \) can be calculated from the RMS current using the relationship: \[ I_{peak} = I_{RMS} \cdot \sqrt{2} \] Substituting the value of \( I_{RMS} \): \[ I_{peak} = 10 \cdot \sqrt{2} \approx 14.14 \, A \] ### Final Answer The peak value of current is approximately \( 14.14 \, A \). ---
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