To solve for the voltage drop across the 4.0 µF capacitor (C₁) when two capacitors are connected in series, we can use the following steps:

1. Find the equivalent capacitance of the series combination:

When capacitors are connected in series, the total or equivalent capacitance $C_{\text{eq}}$ is found using the formula:

$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$

Given:

• $C_1 = 4.0 \, \mu F = 4.0 \times 10^{-6} \, F$
• $C_2 = 7.0 \, \mu F = 7.0 \times 10^{-6} \, F$

Now, substitute the values:

$\frac{1}{C_{\text{eq}}} = \frac{1}{4.0 \times 10^{-6}} + \frac{1}{7.0 \times 10^{-6}}$

$\frac{1}{C_{\text{eq}}} = 2.5 \times 10^{5} + 1.42857 \times 10^{5}$

$\frac{1}{C_{\text{eq}}} = 3.92857 \times 10^{5}$

Thus,

$C_{\text{eq}} = \frac{1}{3.92857 \times 10^{5}} = 2.545 \times 10^{-6} \, F = 2.545 \, \mu F$

2. Calculate the total charge (Q) stored on the capacitors:

Since capacitors in series have the same charge, we can calculate the total charge using the total applied voltage and the equivalent capacitance:

$Q = C_{\text{eq}} \cdot V$

where:

• $V = 90.0 \, V$
• $C_{\text{eq}} = 2.545 \times 10^{-6} \, F$

$Q = 2.545 \times 10^{-6} \times 90.0 = 2.2905 \times 10^{-4} \, C$

3. Find the voltage drop across $C_1$ (4.0 µF capacitor):

For a capacitor, the voltage drop $V$ is given by the relation:

$V_1 = \frac{Q}{C_1}$

Substitute the values:

$V_1 = \frac{2.2905 \times 10^{-4}}{4.0 \times 10^{-6}} = 57.26 \, V$

Thus, the voltage drop across the 4.0 µF capacitor is approximately 57.3 V.

Conclusion:

The voltage drop across the 4.0 µF capacitor is 57.3 V.

Would you like further details or have any questions?

Related questions:

1. What is the voltage drop across the 7.0 µF capacitor?
2. How does the voltage change if the capacitors are connected in parallel?
3. What happens to the total capacitance if a third capacitor is added in series?
4. How would the voltage drop change if the 90 V source is replaced with a 120 V source?
5. What is the energy stored in the 4.0 µF capacitor?

Tip:

When capacitors are in series, the same charge flows through each capacitor, but the voltage drop is divided inversely according to their capacitance values.