AP EAMCET · PHYSICS · Capacitance
Three capacitors of capacitances \(C_1=2 \mu \mathrm{F}\), \(C_2=3 \mu \mathrm{F}\) and \(C_3=5 \mu \mathrm{F}\) are connected in series. A potential difference of \(155 \mathrm{~V}\) is applied across the combination. Choose the correct option.
- A Least potential difference is across \(C_3\). Equivalent capacitance of combination is \(\left(\frac{30}{31}\right) \mu \mathrm{F}\). The voltage across \(C_1\) is \(75 \mathrm{~V}\).
- B Least potential difference is across \(C_1\). Equivalent capacitance of combination is \(\left(\frac{30}{51}\right) \mu \mathrm{F}\). The voltage across \(C_2\) is \(50 \mathrm{~V}\).
- C Least potential difference is across \(C_1\). Equivalent capacitance of combination is \(\left(\frac{30}{31}\right) \mu \mathrm{F}\). The voltage across \(C_3\) is \(30 \mathrm{~V}\).
- D Least potential difference is across \(\mathrm{C}_2\). Equivalent capacitance of combination is \(\left(\frac{30}{31}\right) \mu \mathrm{F}\). The voltage across \(C_1\) is \(50 \mathrm{~V}\).
Answer & Solution
Correct Answer
(A) Least potential difference is across \(C_3\). Equivalent capacitance of combination is \(\left(\frac{30}{31}\right) \mu \mathrm{F}\). The voltage across \(C_1\) is \(75 \mathrm{~V}\).
Step-by-step Solution
Detailed explanation
The given situation is shown in the following figure. Equivalent capacitance is given as \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30} \Rightarrow C=\frac{30}{31} \mu \mathrm{F}\) In series combination of capacitances,…
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