Certainly! Let's solve the problem step-by-step.
We are given:
- The capacitance of the first capacitor, [tex]\( C_1 = 20 \,\mu\mathrm{F} \)[/tex]
- The total desired capacitance of the circuit when two capacitors are connected in series, [tex]\( C_{\text{total}} = 12 \,\mu\mathrm{F} \)[/tex]
We need to find the capacitance of the second capacitor, [tex]\( C_2 \)[/tex].
When capacitors are connected in series, the total capacitance [tex]\( C_{\text{total}} \)[/tex] is given by the formula:
[tex]\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \][/tex]
We can rearrange this formula to solve for [tex]\( C_2 \)[/tex].
First, express [tex]\( \frac{1}{C_2} \)[/tex]:
[tex]\[ \frac{1}{C_2} = \frac{1}{C_{\text{total}}} - \frac{1}{C_1} \][/tex]
Substitute [tex]\( C_{\text{total}} = 12 \,\mu\mathrm{F} \)[/tex] and [tex]\( C_1 = 20 \,\mu\mathrm{F} \)[/tex] into the equation:
[tex]\[ \frac{1}{C_2} = \frac{1}{12} - \frac{1}{20} \][/tex]
To proceed, we need a common denominator to combine these fractions. The least common multiple of 12 and 20 is 60. Let's convert these fractions:
[tex]\[ \frac{1}{12} = \frac{5}{60} \][/tex]
[tex]\[ \frac{1}{20} = \frac{3}{60} \][/tex]
Now, perform the subtraction:
[tex]\[ \frac{1}{C_2} = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30} \][/tex]
So, the reciprocal of [tex]\( C_2 \)[/tex] is [tex]\( \frac{1}{30} \)[/tex]. To find [tex]\( C_2 \)[/tex], take the reciprocal of this value:
[tex]\[ C_2 = 30 \,\mu\mathrm{F} \][/tex]
Thus, the value of the second capacitor that needs to be connected in series with the 20 μF capacitor to achieve a total circuit capacitance of 12 μF is:
[tex]\[ C_2 = 30 \,\mu\mathrm{F} \][/tex]
So our final result is:
- The reciprocal calculation: [tex]\( \frac{1}{C_2} = 0.033333333333333326 \)[/tex]
- The capacitance of the second capacitor: [tex]\( C_2 = 30 \,\mu\mathrm{F} \)[/tex]
