MHT CET · Physics · Alternating Current
The capacitive reactance of a capacitor ' \(\mathrm{C}\) ' is \(\mathrm{X} \Omega\). Both, the frequency of a.c. supply and capacitance of the above capacitor are doubled. The new capacitive reactance will be
- A \(\frac{X}{4} \Omega\)
- B \(\frac{\mathrm{X}}{2} \Omega\)
- C \(2 \mathrm{X} \Omega\)
- D \(4 \mathrm{X} \Omega\)
Answer & Solution
Correct Answer
(A) \(\frac{X}{4} \Omega\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{X}_{\mathrm{C}}=\mathrm{X} \Omega\)
\(\Rightarrow \frac{1}{2 \pi \mathrm{fC}}=\mathrm{X} \Omega\)
New Capacitance \(\mathrm{C}=2 \mathrm{C}\) and new frequency
\(\mathrm{f}^{\prime}=2 \mathrm{f}\)
\(\therefore \text {New capacitive reactance } \mathrm{X}_{\mathrm{C}}^{\prime}\) \(=\frac{1}{2 \pi(2 \mathrm{f})(2 \mathrm{C})} \)
\( =\frac{1}{(2 \pi)(4 \mathrm{fC})} \)
\( =\frac{1}{4} \mathrm{X}_{\mathrm{C}} \)
\( =\frac{\mathrm{X}}{4} \Omega\)
\(\Rightarrow \frac{1}{2 \pi \mathrm{fC}}=\mathrm{X} \Omega\)
New Capacitance \(\mathrm{C}=2 \mathrm{C}\) and new frequency
\(\mathrm{f}^{\prime}=2 \mathrm{f}\)
\(\therefore \text {New capacitive reactance } \mathrm{X}_{\mathrm{C}}^{\prime}\) \(=\frac{1}{2 \pi(2 \mathrm{f})(2 \mathrm{C})} \)
\( =\frac{1}{(2 \pi)(4 \mathrm{fC})} \)
\( =\frac{1}{4} \mathrm{X}_{\mathrm{C}} \)
\( =\frac{\mathrm{X}}{4} \Omega\)
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