WBJEE · Physics · Capacitance
Three capacitors of capacitance 1.0,2.0 and \(5.0 \mu \mathrm{F}\) are connected in series to a \(10 \mathrm{V}\) source. The potential difference across the \(2.0 \mu \mathrm{F}\) capacitor is
- A \(\frac{100}{17} \mathrm{V}\)
- B \(\frac{20}{17} \mathrm{v}\)
- C \(\frac{50}{17} \mathrm{V}\)
- D \(10 \mathrm{V}\)
Answer & Solution
Correct Answer
(C) \(\frac{50}{17} \mathrm{V}\)
Step-by-step Solution
Detailed explanation
When the capacitors are connected in series, the resultant capacitance of combination \[ \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \] \[ \begin{array}{l} =\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10} \\ C=\frac{10}{17} \mu \mathrm{F} \end{array} \] The…
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