Question

When an ac source with a peak potential difference of 60 V is connected to a series RLC circuit, the peak current is 2 A. The capacitor is $C = 50 \mu F$, the frequency is 90 Hz, and the current leads the potential difference by $50^\circ$. Find the resistance and the inductance.

Name: When an ac source with a peak potential difference of 60 V is connected to a series RLC circuit, the peak current is 2 A. The capacitor is C = 50 μ F, the frequency is 90 Hz, and the current leads the potential difference by 50^∘. Find the resistance and the inductance.
Uploaded: 2022-07-04T10:08:55-08:00
Duration: 3 min 43 s
Channel: Suman K
Description: When an ac source with a peak potential difference of 60 V is connected to a series RLC circuit, the peak current is 2 A. The capacitor is C = 50 μ F, the frequency is 90 Hz, and the current leads the potential difference by 50^∘. Find the resistance and the inductance.

          When an ac source with a peak potential difference of 60 V is connected to a series RLC circuit, the peak current is 2 A. The capacitor is $C = 50 \mu F$, the frequency is 90 Hz, and the current leads the potential difference by $50^\circ$. Find the resistance and the inductance.

When an ac source with a peak potential difference of 60 V is connected to a series RLC circuit, the peak current is 2 A. The capacitor is C = 50 μ F, the frequency is 90 Hz, and the current leads the potential difference by 50^∘. Find the resistance and the inductance.

Added by Jesus C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Suman K

07/04/2022

Step 1

The formula for angular frequency is ω = 2πf. Given f = 90Hz, ω = 2π(90) = 180π rad/s. Show more…

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When a source with a peak potential difference of 90V is connected to a series RC circuit, the peak current is 2A. The capacitor is C=50μF, the frequency is 90Hz, and the current leads the potential difference by 50°. Find the resistance and the inductance.