In Fig. 25-28, a potential difference V=100 V is applied across a capacitor arrangement with ca

In Fig. 25-28, a potential difference V=100 V is applied...

In Fig. $25-28$, a potential difference $V=100 \mathrm{~V}$ is applied across a capacitor arrangement with capacitances $C_{1}=10.0 \mu \mathrm{F}$, $C_{2}=5.00 \mu \mathrm{F}$, and $C_{3}=4.00 \mu \mathrm{F}$. What are (a) charge $q_{3}$, (b) potential difference $V_{3}$, and $(\mathrm{c})$ stored energy $U_{3}$ for capacitor 3, (d) $q_{1}$, (e) $V_{1}$, and (f) $U_{1}$ for capacitor 1 , and $(\mathrm{g}) q_{2}$, (h) $V_{2}$, and (i) $U_{2}$ for capacitor 2? - $^{\circ}

In Fig. $25-28$, a potential difference $V=100 \mathrm{~V}$ is applied across a capacitor arrangement with capacitances $C_{1}=10.0 \mu \mathrm{F}$, $C_{2}=5.00 \mu \mathrm{F}$, and $C_{3}=4.00 \mu \mathrm{F}$. What are (a) charge $q_{3}$, (b) potential difference $V_{3}$, and $(\mathrm{c})$ stored energy $U_{3}$ for capacitor 3, (d) $q_{1}$,
(e) $V_{1}$, and (f) $U_{1}$ for capacitor 1 , and $(\mathrm{g}) q_{2}$, (h) $V_{2}$, and (i) $U_{2}$ for capacitor 2?
- $^{\circ}

Key Concepts

Capacitance

Capacitance is a measure of a capacitor’s ability to store electrical charge per unit potential difference. It is defined by the relation q = C × V, where q is the charge stored, C is the capacitance, and V is the voltage applied. This concept is fundamental in understanding how capacitors operate in circuits.

Series and Parallel Capacitor Combinations

When capacitors are connected in series, they all carry the same charge although the voltage divides among them, whereas in a parallel arrangement, each capacitor experiences the same voltage while the charge distributes according to their capacitance. Analyzing these configurations requires understanding these distinctive behaviors, which is crucial for accurately determining charge, voltage, and stored energy in each capacitor.

Charge-Voltage Relationship in Capacitors

This key concept involves the direct relationship between charge, capacitance, and voltage, encapsulated in the formula q = CV. It is essential for calculating the individual charges or voltages on capacitors when the configuration (series or parallel) and the applied potential difference are known.

Energy Stored in a Capacitor

The energy stored within a capacitor can be determined using the formula U = ½ CV² or alternatively U = q²/(2C). This concept enables the evaluation of how much energy is held in a capacitor, which is vital in energy storage applications and in understanding the dynamics of transient circuits.

Transcript

00:01 In this problem, we're considering three capacitors connected to a potential difference as shown in the circuit diagram, and we want to find the charge, potential difference, and energy stored for each of these three capacitors.

00:19 So in order to do this, we're going to rely pretty heavily on the formula q equals cv, which will give the charge on a capacitor if you know its capacitance and.

00:30 Voltage and then u equals one half c v squared which gives the potential energy stored in a capacitor again if you know it's capacitance and voltage so for part a let's start with q3 that's going to be capacitance c3 and in this case the voltage across the third capacitor will be equal to the voltage drop v.

01:02 So when you plug that in, you just get that the charge here is 400 microcoules.

01:12 Now we want to find the voltage drop.

01:15 We already said that v3 is equal to v, which was 100 volts.

01:22 And for part c, we now want to find the potential energy u.

01:28 That's going to be one half c3 times v.

01:34 Squared and again you can just go ahead and plug in the numbers for that and you would get two times 10 to the negative 2 jules for part d we want to figure out the charge q1 on capacitor 1 and in order to do that we want to recognize that the charge on capacitor 1 will be equivalent to the charge on capacitor two because they are connected in series and that's going to be equal to the charge queue that can be found by using the equivalent capacitance of those two capacitors times the voltage drop across both capacitors so the equivalent capacitance for capacitors one and two because they are connected in series is one over c1 plus one over c2 and the inverse of that.

02:36 And you can now go ahead and plug in and get that the charge on capacitor one should be 3 .3 times 10 to the negative 4 coulalms.

02:52 Now that we know the charge though, it's pretty easy to find the voltage v1.

02:58 It's going to be the charge q1 divided by the capacitance c1.

03:03 So using that charge that we found in part d...

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