KCET · Physics · Capacitance
How many \(6 \mu \mathrm{F}, 200 \mathrm{~V}\) condensers are needed to make a condenser of \(18 \mu \mathrm{F}, 600 \mathrm{~V}\) ?
- A 9
- B 18
- C 3
- D 27
Answer & Solution
Correct Answer
(D) 27
Step-by-step Solution
Detailed explanation
In series arrangement potential difference is the sum of the individual potential difference of each capacitor.
\(\mathrm{V} =\mathrm{V}_{1}+\mathrm{V}_{2}+\mathrm{V}_{3}+\ldots \)
\(600 =\mathrm{x} \times 200 \)
\(\mathrm{x} =3\)
So, there should be 3 capacitors in series to obtain the required potential difference.
The equivalent capacitance of the 3 capacitors in series is
\(\frac{1}{\mathrm{C}_{\mathrm{eq}}} =\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \)
\(\mathrm{C}_{\mathrm{eq}} =2\)
Now, we require \(18 \mu \mathrm{F}\) capacitance, for this we need 9 such combinations in parallel. Hence, \(\quad 9 \times 3=27\)
\(\mathrm{V} =\mathrm{V}_{1}+\mathrm{V}_{2}+\mathrm{V}_{3}+\ldots \)
\(600 =\mathrm{x} \times 200 \)
\(\mathrm{x} =3\)
So, there should be 3 capacitors in series to obtain the required potential difference.
The equivalent capacitance of the 3 capacitors in series is
\(\frac{1}{\mathrm{C}_{\mathrm{eq}}} =\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \)
\(\mathrm{C}_{\mathrm{eq}} =2\)
Now, we require \(18 \mu \mathrm{F}\) capacitance, for this we need 9 such combinations in parallel. Hence, \(\quad 9 \times 3=27\)
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