Question

The potential difference across two capacitors in series is \(120 . \mathrm{V}\). 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: Calculate the total capacitance of the pair of capacitors, the charge on each capacitor, the potential difference across each capacitor, and the total energy stored by the capacitors, given a potential difference of 120V and capacitances \(C_1 = 1.00 \cdot 10^3 \mu F\) and \(C_2 = 1.50 \cdot 10^3 \mu F\). Answer: The total capacitance is 600 µF. The charge on each capacitor is 72 mC. The potential difference across \(C_1\) is 72V and across \(C_2\) is 48V. The total energy stored by the capacitors is 4.968 Joules.

Step-by-step solution

1. Calculate total capacitance

To determine the total capacitance of the two capacitors in series, we use the formula: $$\frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2}$$ Where \(C_1 = 1.00 \cdot 10^{3} \mu F\) and \(C_2 = 1.50 \cdot 10^{3} \mu F\). Solve for \(C_t\): $$C_t=\frac{1}{\frac{1}{1.00 \cdot 10^{3} \mu F}+\frac{1}{1.50 \cdot 10^{3} \mu F}}=600\mu F$$

2. Calculate charge on each capacitor

The charge is equal in capacitors connected in series. Therefore, we only need to find the charge on one capacitor: $$Q = C_t \cdot V$$ Where \(V = 120V\) and \(C_t = 600\mu F\). Solve for \(Q\): $$Q = 600\mu F \cdot 120 V = 72 mC$$ So, the charge on each capacitor is 72 mC.

3. Calculate potential difference across each capacitor

To find the potential difference across each capacitor, use the equation: $$V_i = \frac{Q}{C_i}\; \text{,where}\; i=1,2$$ For \(C_1 = 1.00 \cdot 10^{3} \mu F\) and \(C_2 = 1.50 \cdot 10^{3} \mu F\): $$V_1 = \frac{72 mC}{1.00 \cdot 10^{3} \mu F} = 72V$$ $$V_2 = \frac{72 mC}{1.50 \cdot 10^{3} \mu F} = 48V$$ The potential difference across \(C_1\) is 72V and across \(C_2\) is 48V.

4. Calculate total energy stored by the capacitors

To determine the total energy stored, we can sum up the energy stored in each capacitor: $$E = E_1 + E_2 = \frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2$$ With \(C_1 = 1.00 \cdot 10^{3} \mu F\), \(C_2 = 1.50 \cdot 10^{3} \mu F\), \(V_1 = 72V\) and \(V_2 = 48V\), we have: $$E = \frac{1}{2}(1.00 \cdot 10^{3} \mu F)(72V)^2 + \frac{1}{2}(1.50 \cdot 10^{3} \mu F)(48V)^2 = 3.24J + 1.728J = 4.968J$$ The total energy stored by the capacitors is 4.968 Joules.

Key concepts

Total Capacitance

When capacitors are connected in series, the total capacitance (C_t) is less than any individual capacitance in the circuit. This is different from how resistors behave in series, where the resistances add up. The formula to find the total capacitance in a series circuit is:

• \( \frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \)

The reason for this formula is that in a series configuration, the charge has only one path to follow through each capacitor. This means they share the charge but the potential difference divides among them.
For the capacitors in our exercise:

• \( C_1 = 1.00 \times 10^3 \, \mu F \)
• \( C_2 = 1.50 \times 10^3 \, \mu F \)
• \( \frac{1}{C_t} = \frac{1}{1000} + \frac{1}{1500} \)

Calculating yields a total capacitance of 600 microfarads (\( \mu F \)). This is an important step because the total capacitance determines how the capacitors interact collectively in a circuit.

Charge on Capacitors

In a series circuit, each capacitor stores an equal amount of electrical charge (\(Q\)). This element of physics is helpful for calculating the charge without needing to do individual measurements for each capacitor.The charge on the capacitors can be calculated using the total capacitance and the total voltage supplied to the circuit with the formula:

• \( Q = C_t \cdot V \)

In our case:

• Total capacitance (\(C_t\)) is 600 microfarads
• The potential difference (\(V\)) across the series is 120 volts
• Substitute to get: \( Q = 600 \, \mu F \times 120 \, V = 72 \, mC \)

The charge on each capacitor is therefore 72 milliocoulombs (\( mC \)). Even though the individual potential differences across each capacitor differ, the charge remains uniform.

Potential Difference

The potential difference (voltage) across each capacitor in a series is not the same. This is because, in a series circuit, the total voltage applied (120V, in this case) divides among the capacitors based on their individual capacitances.To find the voltage across each capacitor, use the formula:

• \( V_i = \frac{Q}{C_i} \)

Here's how:

• For \( C_1 = 1.00 \times 10^3 \, \mu F \), \( V_1 = \frac{72 \, mC}{1000 \, \mu F} = 72 V \)
• For \( C_2 = 1.50 \times 10^3 \, \mu F \), \( V_2 = \frac{72 \, mC}{1500 \, \mu F} = 48 V \)

Thus, the first capacitor has a potential difference of 72 volts, while the second has 48 volts. Remember, the sum of these individual voltages equals the total voltage supplied across the series (120V).

Energy Stored in Capacitors

Capacitors store energy in the form of an electrostatic field, and the total energy stored can be calculated based on the individual capacitance and the potential difference across each capacitor.The energy (E) stored in a capacitor is given by:

• \( E = \frac{1}{2} C V^2 \)

So for our capacitors:

• For \( C_1 \): \( E_1 = \frac{1}{2} \times 1.00 \times 10^3 \, \mu F \times (72 \, V)^2 = 3.24 \, J \)
• For \( C_2 \): \( E_2 = \frac{1}{2} \times 1.50 \times 10^3 \, \mu F \times (48 \, V)^2 = 1.728 \, J \)

Therefore, the total energy stored in the series capacitors is the sum \( 3.24 \, J + 1.728 \, J = 4.968 \, J \).
This check ensures that capacitors store and release energy correctly for use in electronic circuits.