Question

1) (20 marks) Using superposition, for the network of the figure below, determine the voltage $V_L$ ($\mu = 20$). Use the superposition approach. $V = 2 V \angle 0^\circ$ $I = 2 mA \angle 0^\circ$ $R_1 = 5 k\Omega$ $X_C = 1 k\Omega$ $R_2 = 4 k\Omega$

Name: 1 20 marks using superposition for the network of the figure below determine the voltage vl mu 20 use the superposition approach v 2 v angle 0circ i 2 ma angle 0circ r1 5 komega xc 1 kome 95872
Uploaded: 2020-06-26T15:37:49-08:00
Duration: 3 min 19 s
Channel: Tanishq Gupta
Description: 1 20 marks using superposition for the network of the figure below determine the voltage vl mu 20 use the superposition approach v 2 v angle 0circ i 2 ma angle 0circ r1 5 komega xc 1 kome 95872

          1) (20 marks) Using superposition, for the network of the figure below, determine the voltage $V_L$ ($\mu = 20$). Use the superposition approach.
$V = 2 V \angle 0^\circ$
$I = 2 mA \angle 0^\circ$
$R_1 = 5 k\Omega$
$X_C = 1 k\Omega$
$R_2 = 4 k\Omega$

1 20 marks using superposition for the network of the figure below determine the voltage vl mu 20 use the superposition approach v 2 v angle 0circ i 2 ma angle 0circ r1 5 komega xc 1 kome 95872

Added by Tom-S N.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Tanishq Gupta

06/26/2020

Step 1

The circuit becomes a voltage divider. $V_{L1} = V \frac{R_2 \parallel X_C}{R_1 + R_2 \parallel X_C} = 2 \frac{4 \times 1}{5 + \frac{4 \times 1}{4+1}} = 2 \frac{4}{5 + 0.8} = 2 \frac{4}{5.8} = 1.38 V$ Show more…

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1) (20 marks) Using superposition, for the network of the figure below, determine the voltage $V_L$ ($\mu = 20$). Use the superposition approach. $V = 2 V \angle 0^\circ$ $I = 2 mA \angle 0^\circ$ $R_1 = 5 k\Omega$ $X_C = 1 k\Omega$ $R_2 = 4 k\Omega$