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Chapter 24: Problem 44

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

Expert verified
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

01

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 \(C_t\): $$C_t=\frac{1}{\frac{1}{1.00 \cdot 10^{3} \mu F}+\frac{1}{1.50 \cdot 10^{3} \mu F}}=600\mu F$$
02

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.
03

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.
04

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.

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