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In (Figure 1), each capacitance C₁ is 6.6 μF, and each capacitance C₂ is 4.4 μF. Figure C₁ C₂= HH C₁ C₁ < 1 of 1 > HE C₁ C₁ C₂= C₁ HH =C C₁ Part A Compute the equivalent capacitance of the network between points a and b. Express your answer in farads. Ceq 2.2x10-6 F Submit ✓ Correct Part B Previous Answers Compute the charge on the capacitor C₁ nearest to a when Val = 500 V. Express your answer in coulombs. Q=1.1×10-3 C Submit Previous Answers

In (Figure 1), each capacitance C₁ is 6.6 μF, and each capacitance C₂ is 4.4 μF. Figure C₁ C₂= HH C₁ C₁ < 1 of 1 > HE C₁ C₁ C₂= C₁ HH =C C₁ Part A Compute the equivalent capacitance of the network between points a and b. Express your answer in farads. Ceq 2.2x10-6 F Submit ✓ Correct Part B Previous Answers Compute the charge on the capacitor C₁ nearest to a when Val = 500 V. Express your answer in coulombs. Q=1.1×10-3 C Submit Previous Answers

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Hello there again. To start off, I AM BLIND AND CAN NOT READ IMAGES YOU MAY ATTEMPT TO RESPOND WITH. Please do not do that,  it wastes your time and mine and would just like it written out in text like what you are litterly reading right now. Thank you for understand. will give great  possitive  remarks.

 

So the question I am working on  I have 1 part left and am not sure on how to calculate the final part. 

THE DESCRIPTION TELLS ME :

 

A CIRCUIT INCLUDES 9 capacitors. The circuit has end points A and B. Capacitor C  Capacitor C_1  is next to point A. Another capacitor c_1 is  next to point B. Between these two capacitors, capacitor c_2 is connected parallel with the following combination.  capacitor c_1 is between c_2 and point C. on the other side of c_2, another capacitor c_1 is between c_2 and point D.There are  two parallel branches  between points C and D. There is one more capacitor c_2 on the first branch, and there are three capacitors c_1 in series on the second branch.

 

that is it 

The question goes:

In (Figure 1), each capacitance C1C1 is 6.6 μFμF, and each capacitance C2C2 is 4.4 μFμF.

 

Figure

1 of 1

 

Part A

Part complete

Compute the equivalent capacitance of the network between points aa and bb.

Express your answer in farads.

CeqCeq =

2.2×10−6

FF

Got this one correct.

 

Part B

Part complete

Compute the charge on the capacitor C1C1 nearest to aa when VabVab = 500 VV.

Express your answer in coulombs.

Qa1Q1a =

1.1×10−3

CC

This one is correct to. 

Part C

Part complete

Compute the charge on the capacitor C1C1 nearest to bb when VabVab = 500 VV.

Express your answer in coulombs.

Qb1Q1b =

1.1×10−3

CC

 

THIS ONE WAS CORRECT TOO.

 

Part D

Part complete

Compute the charge on the capacitor C2C2 nearest to aa and bb when VabVab = 500 VV.

Express your answer in coulombs.

Q2Q2 =

7.3×10−4

CC

 

THIS ONE I GOT CORRECT TOO AND YOU DON'T NEED TO WASTE YOUR TIME ON THE ABOVE PARTS.

 

The final part goes:

Part E

With 500 VV across aa and bb, compute VcdVcd.

Express your answer in volts.

 

If you could direct me with normal text and not images on how to answer this final question 

i would be very appericative and will throw possitive responses back :). 

 

Thanks again.

 

Also will attempt to give you a image but no gaurentee and have the description above too.

 

 

 

 

 

 

 

 

**CH 24: Problem 24.59** In Figure 1, each capacitance \( C_1 \) is 6.6 µF, and each capacitance \( C_2 \) is 4.4 µF. **Diagram Description:** The circuit diagram shows a network of capacitors between points \( a \) and \( b \). The capacitors are arranged as follows: - Three \( C_1 \) capacitors are connected in series, labeled \( a \) to \( c \). - Two \( C_2 \) capacitors are connected in parallel with each other and are in between \( C_1 \) capacitors from nodes \( c \) to \( d \). - At point \( b \), two more \( C_1 \) capacitors are connected in parallel to complete the circuit. **Part A:** Compute the equivalent capacitance of the network between points \( a \) and \( b \). - Express your answer in farads. \[ C_{\text{eq}} = 2.2 \times 10^{-6} \, \text{F} \] - Status: Correct **Part B:** Compute the charge on the capacitor \( C_1 \) nearest to \( a \) when \( V_{ab} = 500 \, \text{V} \). - Express your answer in coulombs. \[ Q_{1}^{a} = 1.1 \times 10^{-3} \, \text{C} \] - Status: Correct
Transcribed Image Text:**CH 24: Problem 24.59** In Figure 1, each capacitance \( C_1 \) is 6.6 µF, and each capacitance \( C_2 \) is 4.4 µF. **Diagram Description:** The circuit diagram shows a network of capacitors between points \( a \) and \( b \). The capacitors are arranged as follows: - Three \( C_1 \) capacitors are connected in series, labeled \( a \) to \( c \). - Two \( C_2 \) capacitors are connected in parallel with each other and are in between \( C_1 \) capacitors from nodes \( c \) to \( d \). - At point \( b \), two more \( C_1 \) capacitors are connected in parallel to complete the circuit. **Part A:** Compute the equivalent capacitance of the network between points \( a \) and \( b \). - Express your answer in farads. \[ C_{\text{eq}} = 2.2 \times 10^{-6} \, \text{F} \] - Status: Correct **Part B:** Compute the charge on the capacitor \( C_1 \) nearest to \( a \) when \( V_{ab} = 500 \, \text{V} \). - Express your answer in coulombs. \[ Q_{1}^{a} = 1.1 \times 10^{-3} \, \text{C} \] - Status: Correct
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