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Chapter 11: Problem 1627

There are 10 condensers each of capacity \(5 \mu \mathrm{F}\). The ratio between maximum and minimum capacities obtained from these condensers will be (A) \(40: 1\) (B) \(25: 5\) (C) \(60: 3\) (D) \(100: 1\)

Short Answer

Expert verified
The maximum capacity is achieved by connecting all the condensers in parallel, resulting in \(50 \mu F\). The minimum capacity is obtained by connecting all the condensers in series, resulting in \(\frac{1}{2} \mu F\). The ratio between these capacities is \(\frac{50 \mu F}{\frac{1}{2} \mu F} = 100:1\), making the answer option (D).

Step by step solution

01

Find the maximum capacity

To find the maximum capacity, we can connect all the condensers in parallel. In this case, the total capacity is the sum of the individual capacities. As each condenser has a capacity of \(5 \mu F\), the total capacity for all 10 condensers connected in parallel will be: \(C_{max} = 10 \times 5 \mu F = 50 \mu F\)
02

Find the minimum capacity

To find the minimum capacity, we can connect all the condensers in series. In this case, the reciprocal of the total capacity is equal to the sum of the reciprocals of the individual capacities. Thus, \(\frac{1}{C_{min}} = \sum\limits^{10}_{i=1} \frac{1}{C_i}\) As each condenser has a capacity of \(5 \mu F\), we can write: \(\frac{1}{C_{min}} = 10 \times \frac{1}{5 \mu F} = \frac{10}{5 \mu F}\) Now, we can find the minimum capacity by taking the reciprocal of the previous expression: \(C_{min} = \frac{1}{\frac{10}{5 \mu F}} = \frac{5 \mu F}{10} = \frac{1}{2} \mu F\)
03

Find the ratio between maximum and minimum capacities

Now that we have the maximum and minimum capacities, we can determine the ratio between them: Ratio = \(\frac{C_{max}}{C_{min}} = \frac{50 \mu F}{\frac{1}{2} \mu F} = 100 \) Thus, the ratio between maximum and minimum capacities obtained from these condensers is 100:1, which corresponds to option (D).

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