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2 years ago

5

Find the current in 1 the electrical circuit below

1 answer:

Lesechka [4]2 years ago

3 0

Answer:

c. 25 μA

Explanation:

The dependent current source means that 31 times i1 will flow through the 6kΩ resistor, effectively increasing its voltage drop to 31 times that which would result from i1 alone. In effect, the 6kΩ resistor behaves in the left-side circuit loop as though it were 31×6kΩ = 186kΩ (with no dependent current source).

Then the current i1 is equivalent to that created by a 5+1 = 6V source through a 54kΩ +186kΩ = 240kΩ circuit impedance.

(6V)/(240kΩ) = 25 μA

_____

<em>Additional comment</em>

The voltage across the 6kΩ resistor is (186/240)·6V = 4.65V, and the 25 μA current generates a voltage of 30·(25 μA)(1.8kΩ) = 1.35V across the 1.8kΩ resistor. This means the voltage source at the right side of the diagram needs to be at least 4.65 +1.35 = 6.0V in order to support the calculated voltage drops.

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Dima020 [189]

Answer:

248.756 mV

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Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

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The Thevenin equivalent source at the other terminal of the bridge is ...

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The open-circuit voltage is the difference between these terminal voltages:

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The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0

2 years ago

a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

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$R =\frac{1}{2*3.14*100*10^-^9*8000}$

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b) diagram is attached

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$|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}$

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=20log0.923

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