Question

Calculate the voltage across C1 in the following circuit if the total voltage is 12 V. Assume C1 = 8.0 μF, C2 = 11 μF.

Answer 1

To calculate the voltage across capacitor C1, we need to consider the charges on each capacitor in the circuit. The charge on C2 and C3 is 16.0 μC and 32.0 μC, respectively. The voltage across C1 is 2.0 V.

Step-by-step explanation:

To calculate the voltage across capacitor C1, we need to consider the network of capacitors in the circuit. Using the information given, we can determine the charges on each capacitor by multiplying its capacitance by the potential difference across it. In this case, the potential difference across capacitors C2 and C3 is 8.0 V. Therefore, the charge on C2 is 16.0 μC and the charge on C3 is 32.0 μC.

Since C1 is in series with C2 and C3, the charges on C2 and C3 are also the charges on C1. Thus, the charge on C1 is 16.0 μC.

The voltage across C1 can be calculated using the formula V = Q/C, where Q is the charge on C1 and C is the capacitance of C1. Given that C1 = 8.0 μF and Q1 = 16.0 μC, the voltage across C1 is:

V = Q/C = (16.0 μC)/(8.0 μF) = 2.0 V.

Answer 2

The charge Q on each capacitor is approximately
`$55.58 \mu C$` (microcoulombs). Using this charge, the voltage across capacitor
`$C_1\left(V_(C 1)\right)$` is approximately
`$6.95 \mathrm{~V}$`.

To calculate the voltage across
`$C_1$` in the circuit you've provided, where
`$C_1$` and
`$C_2$` are in series, we'll use the formula for capacitors in series to find the equivalent capacitance,
`$C_{\mathrm{eq}}$`, and then use that to find the voltage across
`$C_1$`.

The formula for capacitors in series is:

`$\frac{1}{C_{\mathrm{eq}}}=(1)/(C_1)+(1)/(C_2)$`

Solving for
`$C_{\text {eq }}$` gives us:

`$C_{\mathrm{eq}}=(1)/((1)/(C_1)+(1)/(C_2))$`

Once we have
`$C_{\mathrm{eq}}$`, we can use the voltage division rule, which is analogous to the current division rule in resistors. The voltage across
`$C_1, V_(C 1)$`, can be found using:

`$V_(C 1)=V_{\text {total }} * \frac{C_{\text {eq }}}{C_1}$`

Given:

• Total voltage,
  `$V_{\text {total }}=12 \mathrm{~V}$`
• Capacitance of
  `$C_1, C_1=8.0 \mu F$`
• Capacitance of
  `$C_2, C_2=11 \mu \mathrm{F}$`

The total voltage in a series circuit divides over the capacitors in proportion to their capacitances. Since the charge Q on each capacitor is the same, we have:

`$\begin{aligned} & Q=C_1 V_(C 1)=C_2 V_(C 2) \\ & V_(C 1)=(Q)/(C_1), V_(C 2)=(Q)/(C_2)\end{aligned}$`

And we know that:

`$$V_{\text {total }}=V_(C 1)+V_(C 2)$$`

So we have:

`$$V_{\text {total }}=(Q)/(C_1)+(Q)/(C_2)$$`

We can solve for $Q$ using:

`$$Q=V_{\text {total }} * (C_1 * C_2)/(C_1+C_2)$$`

Then, we can find $V_{C 1}$ with:

`$$V_(C 1)=(Q)/(C_1)$$`

Here's the step-by-step calculation:

1. Calculate the total charge $Q$ shared by both capacitors:

`$$Q=V_{\text {total }} * (C_1 \cdot C_2)/(C_1+C_2)=12 \cdot (8.0 * 10^(-6) \cdot 11 * 10^(-6))/(8.0 * 10^(-6)+11 * 10^(-6)) \approx 55.58 \mu C$$`

2. Calculate the voltage across $C_1$ using the charge $Q$ :

`$$V_(C 1)=(Q)/(C_1)=(55.58 * 10^(-6))/(8.0 * 10^(-6)) \approx 6.95 \mathrm{~V}$$`

So, the voltage across capacitor
`$C_1$` in your circuit is approximately 6.95 volts.

The complete question is here:

Calculate the voltage across C1 in the following circuit if the total voltage is 12 V. Assume C1 = 8.0 μF, C2 = 11 μF.