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Chapter 11: Problem 1622

The capacitors of capacitance \(4 \mu \mathrm{F}, 6 \mu \mathrm{F}\) and $12 \mu \mathrm{F}$ are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases? (A) \(2: 3\) (B) \(11: 1\) (C) \(1: 11\) (D) \(1: 3\)

Short Answer

Expert verified
The ratio of equivalent capacitance in the two cases is \(1: 11\) (Option C).

Step by step solution

01

To find the equivalent capacitance (\(C_s\)) in a series configuration, we use the following formula: \[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\] where \(C_1 = 4 \mu F\), \(C_2 = 6 \mu F\), and \(C_3 = 12 \mu F\). #Step 2: Substitute values and solve for equivalent capacitance in series#

Substitute the values of the capacitors and solve for \(C_s\): \[\frac{1}{C_s} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12}\] #Step 3: Simplify the equation for Cs#
02

To simplify the equation, we can find the least common denominator of \(4\), \(6\), and \(12\), as follows: \[\frac{1}{C_s} = \frac{3+2+1}{12}\] #Step 4: Solve for Cs#

Solve for \(C_s\): \[\frac{1}{C_s} = \frac{6}{12}\] \[C_s = 2 \mu F\] #Step 5: Calculate the equivalent capacitance in parallel configuration#
03

To find the equivalent capacitance (\(C_p\)) in a parallel configuration, we use the following formula: \[C_p = C_1 + C_2 + C_3\] #Step 6: Substitute values and solve for equivalent capacitance in parallel#

Substitute the values of the capacitors and solve for \(C_p\): \[C_p = 4 + 6 + 12\] \[C_p = 22 \mu F\] #Step 7: Calculate the ratio of equivalent capacitance in series and parallel configurations#
04

Now that we have the equivalent capacitance in series and parallel configurations, we can find the ratio between these values: \[\text{Ratio}= \frac{C_s}{C_p}\] #Step 8: Substitute values and solve for ratio#

Substitute the values of \(C_s\) and \(C_p\) and solve for the ratio: \[\text{Ratio}= \frac{2}{22}\] \[\text{Ratio}= \frac{1}{11}\] Therefore, the ratio of equivalent capacitance in the two cases is \(\boldsymbol{1: 11}\) (Option C).

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Most popular questions from this chapter

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