Question

The potential difference across two capacitors in series is \(120 .\) V. The capacitances are \(C_{1}=1.00 \cdot 10^{3} \mu \mathrm{F}\) and \(C_{2}=1.50 \cdot 10^{3} \mu \mathrm{F}\). a) What is the total capacitance of this pair of capacitors? b) What is the charge on each capacitor? c) What is the potential difference across each capacitor? d) What is the total energy stored by the capacitors?

Short answer

Question: Determine the following information for two capacitors in series, with capacitances \(C_1 = 1.0 \cdot 10^3\mu F\) and \(C_2 = 1.5 \cdot 10^3\mu F\), and a potential difference of 120 V across the pair: a) The total capacitance of the pair of capacitors b) The charge on each capacitor c) The potential difference across each capacitor d) The total energy stored by the capacitors Answer: a) The total capacitance of the pair of capacitors is 600 \(\mu\)F. b) The charge on each capacitor is 72 mC. c) The potential difference across \(C_1\) is 72 V and across \(C_2\) is 48 V. d) The total energy stored by the capacitors is 4.84704 J.

Step-by-step solution

Find the total capacitance of capacitors in series

For capacitors in series, the total capacitance is given by: \(\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2}\) Given \(C_1 = 1.0 \cdot 10^3\mu F\) and \(C_2 = 1.5 \cdot 10^3\mu F\), we can plug these values into the equation and solve for \(C_{total}\).

Calculate the total capacitance

Using the formula from Step 1, we can calculate \(C_{total}\) as follows: \(\frac{1}{C_{total}} = \frac{1}{1.0 \cdot 10^3\mu F} + \frac{1}{1.5 \cdot 10^3\mu F}\) To find \(C_{total}\), take the inverse of both sides: \(C_{total} = \frac{1}{\frac{1}{1.0 \cdot 10^3\mu F} + \frac{1}{1.5 \cdot 10^3\mu F}}\) \(C_{total} = 600 \mu F\) #a) Total capacitance of the pair of capacitors is 600 \(\mu\)F.

Calculate the charge on each capacitor

Since they are in series, both have the same charge \(Q\). Using the formula \(Q = C\Delta V\) for capacitors, we have: \(Q = C_{total} \Delta V\) Given \(\Delta V = 120\) V, we can substitute and solve for \(Q\): \(Q = 600\mu F \cdot 120 V = 72 mC\). #b) The charge on each capacitor is 72 mC.

Calculate the potential difference across each capacitor

We have the charge on each capacitor, so we can use \(Q = C\Delta V\) to find the potential difference across each capacitor. For \(C_1\): \(\Delta V_1 = \frac{Q}{C_1} = \frac{72 mC}{1.0 \cdot 10^3 \mu F} = 72 V\) For \(C_2\): \(\Delta V_2 = \frac{Q}{C_2} = \frac{72 mC}{1.5 \cdot 10^3 \mu F} = 48 V\) #c) The potential difference across \(C_1\) is 72 V and across \(C_2\) is 48 V.

Calculate the total energy stored by the capacitors

The energy stored in each capacitor can be calculated using the formula \(U = \frac{1}{2}C\Delta V^2\). We can then sum the energies of the two capacitors to find the total energy. For \(C_1\): \(U_1 = \frac{1}{2}(1.0 \cdot 10^3 \mu F)(72 V)^2 = 3.1104 J\) For \(C_2\): \(U_2 = \frac{1}{2}(1.5 \cdot 10^3 \mu F)(48 V)^2 = 1.73664 J\) Now, sum the energies: \(U_{total} = U_1 + U_2 = 3.1104 J + 1.73664 J = 4.84704 J\) #d) The total energy stored by the capacitors is 4.84704 J.

Key concepts

Total Capacitance Calculation

To understand how capacitors behave when they are combined in series, it's crucial to grasp the concept of total capacitance calculation. Imagine you have several storage containers connected in a single line, each capable of holding a certain amount of water. Here, the total storage capacity is limited by the smallest container, since water can't flow beyond its capacity. Similarly, when capacitors are connected in series, the total capacitance is determined by the formula:
\[ \frac{1}{C_{total}} = \sum_{i=1}^{n} \frac{1}{C_i} \
where \( C_{total} \) represents the total capacitance of the series combination, and \(C_i\) represents the capacitance of each individual capacitor. For two capacitors in series with capacitances \(C_1\) and \(C_2\), we calculate the total capacitance as \( 600 \mu F \) which indicates the overall ability of the series combination to store charge is equivalent to a single capacitor of \( 600 \mu F \). This calculation is vital for designing circuits that require a precise capacitance value.

Charge on Capacitors

One might wonder, how does charge behave in a series of capacitors? When capacitors are connected in series, they all store the same amount of charge, regardless of their individual capacitances. This is akin to filling a series of connected balloons with air; regardless of their size, each balloon ends up with the same amount of air inside.
The charge \(Q\) stored on each capacitor is given by the equation:
\[ Q = C_{total} \Delta V \
where \(\Delta V\) is the total potential difference applied across the series combination. If we connect our pair of capacitors to a \(120\) V battery, each capacitor holds a charge of \(72 mC\). This shared charge is a fundamental property that distinguishes series circuits from parallel ones.

Potential Difference Across Capacitors

If we're curious about the potential difference, or voltage, across each individual capacitor in a series, we must remember that it will vary depending on the capacitor's capacitance. The potential difference across each capacitor in a series is given by the charge on the capacitor divided by its capacitance, expressed as \(\Delta V = \frac{Q}{C}\).
For our two capacitors with charge \(72 mC\), we find that \(C_1\) has a potential difference of \(72 V\) and \(C_2\) has a potential difference of \(48 V\). Think of this as pressure in water pipes; the larger the container (capacitor), the less pressure (voltage) it takes to fill it with a certain amount of water (charge). Understanding the distribution of voltage is crucial for troubleshooting and designing electronic circuits.

Energy Stored in Capacitors

You may be asking yourself, how much energy is actually stored in this series arrangement of capacitors? The energy \(U\) stored in a capacitor can be determined by the formula:
\[ U = \frac{1}{2} C \Delta V^2 \
This formula illustrates that the energy depends not only on the capacitance but also on the square of the potential difference across the capacitor. For our example, the total energy stored in the two capacitors is \(4.84704 J\). This energy storage is akin to the potential energy of water held behind a dam — the higher the water, the greater the potential energy. In the case of capacitors, the 'height' is related to the voltage, and the capacitor's ability to 'hold' the energy is related to its capacitance. Understanding energy storage is essential for applications like flash photography, power stabilization, and many other electronic devices.