Question

Suppose you have the circuit shown below. Initially the switch is open and the capacitors are completely discharged. C1(150?F) C2(50?F) 4.7 k? 5V After the switch has been close for a very long time what is the voltage across the capacitor C2? How long does it take to reach half this value?

          Suppose you have the circuit shown below. Initially the switch is open and
the capacitors are completely discharged.
C1(150?F) C2(50?F)
4.7 k?
5V
After the switch has been close for a very long time what is the voltage
across the capacitor C2?
How long does it take to reach half this value?

Suppose you have the circuit shown below. Initially the switch is open and
the capacitors are completely discharged.
C1(150?F) C2(50?F)
4.7 k?
5V
After the switch has been close for a very long time what is the voltage
across the capacitor C2?
How long does it take to reach half this value?

Added by Susana A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

07/18/2022

Step 1

Given: C1 = 150 pF C2 = 50 pF R = 4.7 kΩ V = 5V Using the formula for capacitors in series, we have: Ceq = C1 * C2 / (C1 + C2) Ceq = 150 * 50 / (150 + 50) Ceq = 37.5 pF The voltage across C2 can be calculated using the voltage divider formula: V2 = V * (C1 / Show more…

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Suppose you have the circuit shown below. Initially, the switch is open and the capacitors are completely discharged. C1 (150 pF) C2 (50 pF) 4.7 k̩ After the switch has been closed for a very long time, what is the voltage across the capacitor C2? How long does it take to reach half this value?