In the circuit shown, the capacitor is ini- tially uncharged. At t₁ = 0, the switch S is moved to position a. R2 www C R1 S b www Vo Find VR₁, the voltage drop across R₁, as a function of time t₁. e-t/ 1. VR₁ = V₁ [1 − e−¹1/(R₁+R₂) C]] 2. VR₁ = Voe (R₁+R2) t₁/(R₁ R₂C) 3. VR₁ =Voet/(R₂C) 4. VR₁ = V₁ [1 − e−¹/(R₁C)] 5. VR₁ = Vo [1-(R₁+Ra) t₁/(R₁ R₂C)] 6. VR₁ = Voet₁/(R₁C) 7. VR₁ = Voet/[(R1+R2) C] 8. VR₁ = Vo [1 − e−1/(R₂C)] Find Vc, the voltage across C, as a function of time t₁. 1. Vc = Voeti/(R₁ C) 2. Vc = Voet/(R2C) 3. Vc = V₁ [1-et/(R₁+R₂) C]] 4. Vc = Voe (R₁+R2) t₁/(R₁ R₂ C) 5. Vc = V₁ [1 − e−(R₁+R₂) t₁/(R₁ R₂ (R₁ R₂ C)] 6. Vc = Voet/[(R₁+R₂)C] 7. Vc Vo 1 e-ti/(R₂ C)] = -e 8. Vc = V₁ [1 − e−/(R₁ C)] -e
In the circuit shown, the capacitor is ini- tially uncharged. At t₁ = 0, the switch S is moved to position a. R2 www C R1 S b www Vo Find VR₁, the voltage drop across R₁, as a function of time t₁. e-t/ 1. VR₁ = V₁ [1 − e−¹1/(R₁+R₂) C]] 2. VR₁ = Voe (R₁+R2) t₁/(R₁ R₂C) 3. VR₁ =Voet/(R₂C) 4. VR₁ = V₁ [1 − e−¹/(R₁C)] 5. VR₁ = Vo [1-(R₁+Ra) t₁/(R₁ R₂C)] 6. VR₁ = Voet₁/(R₁C) 7. VR₁ = Voet/[(R1+R2) C] 8. VR₁ = Vo [1 − e−1/(R₂C)] Find Vc, the voltage across C, as a function of time t₁. 1. Vc = Voeti/(R₁ C) 2. Vc = Voet/(R2C) 3. Vc = V₁ [1-et/(R₁+R₂) C]] 4. Vc = Voe (R₁+R2) t₁/(R₁ R₂ C) 5. Vc = V₁ [1 − e−(R₁+R₂) t₁/(R₁ R₂ (R₁ R₂ C)] 6. Vc = Voet/[(R₁+R₂)C] 7. Vc Vo 1 e-ti/(R₂ C)] = -e 8. Vc = V₁ [1 − e−/(R₁ C)] -e
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![In the circuit shown, the capacitor is ini-
tially uncharged. At t₁ = 0, the switch S is
moved to position a.
R2
www
C
R1
S b
www
Vo
Find VR₁, the voltage drop across R₁, as a
function of time t₁.
e-t/
1. VR₁ = V₁ [1 − e−¹1/(R₁+R₂) C]]
2. VR₁ = Voe (R₁+R2) t₁/(R₁ R₂C)
3. VR₁ =Voet/(R₂C)
4. VR₁ = V₁ [1 − e−¹/(R₁C)]
5. VR₁ = Vo [1-(R₁+Ra) t₁/(R₁ R₂C)]
6. VR₁ = Voet₁/(R₁C)
7. VR₁ = Voet/[(R1+R2) C]
8. VR₁ = Vo [1 − e−1/(R₂C)]
Find Vc, the voltage across C, as a function
of time t₁.
1. Vc = Voeti/(R₁ C)
2. Vc = Voet/(R2C)
3. Vc = V₁ [1-et/(R₁+R₂) C]]
4. Vc = Voe (R₁+R2) t₁/(R₁ R₂ C)
5. Vc = V₁ [1 − e−(R₁+R₂) t₁/(R₁ R₂
(R₁ R₂ C)]
6. Vc = Voet/[(R₁+R₂)C]
7. Vc Vo 1 e-ti/(R₂ C)]
=
-e
8. Vc = V₁ [1 − e−/(R₁ C)]
-e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdaf83079-7816-43d4-9246-9c209cec4377%2Ffae923e1-2c4b-4ce1-850e-2d257e521aca%2Fwp44lj6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In the circuit shown, the capacitor is ini-
tially uncharged. At t₁ = 0, the switch S is
moved to position a.
R2
www
C
R1
S b
www
Vo
Find VR₁, the voltage drop across R₁, as a
function of time t₁.
e-t/
1. VR₁ = V₁ [1 − e−¹1/(R₁+R₂) C]]
2. VR₁ = Voe (R₁+R2) t₁/(R₁ R₂C)
3. VR₁ =Voet/(R₂C)
4. VR₁ = V₁ [1 − e−¹/(R₁C)]
5. VR₁ = Vo [1-(R₁+Ra) t₁/(R₁ R₂C)]
6. VR₁ = Voet₁/(R₁C)
7. VR₁ = Voet/[(R1+R2) C]
8. VR₁ = Vo [1 − e−1/(R₂C)]
Find Vc, the voltage across C, as a function
of time t₁.
1. Vc = Voeti/(R₁ C)
2. Vc = Voet/(R2C)
3. Vc = V₁ [1-et/(R₁+R₂) C]]
4. Vc = Voe (R₁+R2) t₁/(R₁ R₂ C)
5. Vc = V₁ [1 − e−(R₁+R₂) t₁/(R₁ R₂
(R₁ R₂ C)]
6. Vc = Voet/[(R₁+R₂)C]
7. Vc Vo 1 e-ti/(R₂ C)]
=
-e
8. Vc = V₁ [1 − e−/(R₁ C)]
-e
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