Three capacitors of capacitances \(C_1=2 \mu \mathrm{F}\), \(C_2=3 \mu \mathrm{F}\) and \(C_3=5 \mu…
- 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 \(75 \mathrm{~V}\).
- 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 \(50 \mathrm{~V}\).
- 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 \(30 \mathrm{~V}\).
- 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 \(50 \mathrm{~V}\).
Solution
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, charge flowing through each capacitor is same which is given by \(\begin{aligned} q & =C V \\ & =\frac{30}{31} \times 155=150 \mu \mathrm{C}=1.5 \times 10^{-4} \mathrm{C} \end{aligned}\) \(\therefore\) Potential difference across \(C_1\), \(\begin{aligned} V_1 & =\frac{q}{C_1}=\frac{1.5 \times 10^{-4}}{2 \times 10^{-6}}=0.75 \times 10^2 \\ \Rightarrow V_1 & =75 \mathrm{~V} \end{aligned}\) Potential difference across \(C_2\), \(\begin{aligned} V_2 & =\frac{q}{C_2}=\frac{1.5 \times 10^{-4}}{3 \times 10^{-6}}=0.5 \times 10^2 \\ \Rightarrow V_2 & =50 \mathrm{~V} \end{aligned}\) Potential difference across \(C_3\), \(V_3=\frac{q}{C_3}=\frac{1.5 \times 10^{-4}}{5 \times 10^{-6}}=0.3 \times 10^2 \Rightarrow V_3=30 \mathrm{~V}\) Hence, potential difference across \(C_3\) is least as \(30 \mathrm{~V}\).
Asked in: AP EAMCET 2020 (17 Sep Shift 2)