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Please help it’s due very soon and this is my last attempt !!

n (Figure 1)
C
1
= 8.00
mF
,
C
2
= 7.00
mF
,
C
3
= 2.00
mF
, and
C
4
= 5.00
mF
. The capacitor network is connected to an applied potential difference
V
ab
. After the charges on the capacitors have reached their final values, the voltage across
C
3
is 40.0
V

Please help its due very soon and this is my last attempt n Figure 1 C 1 800 mF C 2 700 mF C 3 200 mF and C 4 500 mF The capacitor network is connected to an ap class=

Respuesta :

MathPhys

Answer:

85.9 V

Explanation:

The voltage across C₃ is 40.0 V, so the voltage between a and d must also be 40.0 V.

Find the equivalent capacitance of C₁, C₂, and C₃:

C₁₂₃ = 1 / (1/C₁ + 1/C₂) + C₃

C₁₂₃ = 1 / (1/8.00 + 1/7.00) + 2.00

C₁₂₃ = 5.73 mF

This is in series with C₄.  Capacitors in series have the same charge, so:

Vad C₁₂₃ = Vbd C₄

(40.0) (5.73) = Vbd (5.00)

Vbd = 45.9 V

The total voltage is therefore:

V = Vad + Vbd

V = 40.0 + 45.9

V = 85.9 V

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