Q: A capacitor has a capacitance of $5\text{ }\mu\text{F}$. How much of the charge should be removed in order that the potential difference between its plates decreases by $40\text{ V}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C = 5\text{ }\mu\text{F}$, $\quad \Delta V = 40\text{ V}$, $\quad \Delta Q = ?$
$\begin{aligned}[t] C &= \frac{\Delta Q}{\Delta V} \\ 5 &= \frac{\Delta Q}{40} \\ \Delta Q &= \mathbf{200\text{ }\mu\text{C}} \end{aligned}$
$200\text{ }\mu\text{C}$ of the charge should be removed to decrease potential difference $40\text{ V}$.
Q: When a parallel-plate capacitor is connected to a $50\text{ V}$ battery each plate receives a charge of magnitude $0.002\text{ C}$. Find its capacitance. If the capacitor is connected to a $100\text{ V}$ battery, does the capacitance change? Does the charge still remain the same?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$V = 50\text{ V}$, $\quad Q = 0.002\text{ C}$, $\quad C = ?$
$\begin{aligned}[t] C &= \frac{Q}{V} \\ C &= \frac{0.002}{50} \\ &= 4 \times 10^{-5}\text{ F} \\ &= 40 \times 10^{-6}\text{ F} = \mathbf{40\text{ }\mu\text{F}} \end{aligned}$
Although voltage is increased to $100\text{ V}$, the capacitance does not change but the charge will change.
$V = 100\text{ V}$, $\quad C = 40 \times 10^{-6}\text{ F}$, $\quad Q = ?$
$\begin{aligned}[t] C &= \frac{Q}{V} \\ Q &= C V \\ &= 40 \times 10^{-6} \times 100 \\ &= 40 \times 10^{-4}\text{ C} = \mathbf{0.004\text{ C}} \end{aligned}$
Q: Choose the correct answer from the following.
The plates of a parallel-plate capacitor of capacitance $C$ are brought together to a third of their original separation. The capacitance is now
A. $\frac{1}{9}C$, B. $\frac{1}{3}C$, C. $3C$, D. $9C$.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Original separation $= d$
Original capacitance $= C = \frac{\kappa \varepsilon_0 A}{d}$
New separation $= d' = \frac{1}{3}d$
New capacitance $= C' = \frac{\kappa \varepsilon_0 A}{d'}$
$\begin{aligned}[t]
C' &= \frac{\kappa \varepsilon_0 A}{\frac{1}{3}d} \\
&= 3\frac{\kappa \varepsilon_0 A}{d} \\
&= \mathbf{3\text{ C}}
\end{aligned}$
Q: When the distance between the two parallel-plate of a capacitor is doubled, by what percent does its capacitance change?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Original separation $= d$
Original capacitance $= C = \frac{\kappa \varepsilon_0 A}{d}$
New separation $= d' = 2d$
New capacitance $= C' = \frac{\kappa \varepsilon_0 A}{d'}$
$\begin{aligned}[t]
C' &= \frac{\kappa \varepsilon_0 A}{2d} \\
&= \frac{1}{2} \frac{\kappa \varepsilon_0 A}{d} \\
&= \mathbf{\frac{1}{2}C}
\end{aligned}$
Change in capacitance,
$\Delta C = C - C' = C - \frac{1}{2}C = \mathbf{\frac{1}{2}C}$
$\begin{aligned}[t] \text{percent change} &= \frac{\Delta C}{C} \times 100\% \\ &= \frac{\frac{1}{2}C}{C} \times 100\% \\ &= \mathbf{50\text{ \%}} \end{aligned}$
When the distance between the two parallel-plate of a capacitor is doubled, its capacitance decreases $50\text{ \%}$.
Q: The area of each plate of parallel-plate capacitor is $1\text{ m}^2$ and the distance between two plates is $1\text{ mm}$. If the potential difference the plates is $120\text{ V}$ and the dielectric constant of the material inserted between them is $3$, find (i) the capacitance of the parallel-plate capacitor, (ii) the magnitude of the charge on each plate, (iii) the electric field intensity between the plates.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$A = 1\text{ m}^2$, $d = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$,
$V = 120\text{ V}$, $\kappa = 3$,
$\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}\text{ (or) F m}^{-1}$
(i) $\begin{aligned}[t] C &= \frac{\kappa \varepsilon_0 A}{d} \\ &= \frac{3 \times 8.85 \times 10^{-12} \times 1}{1 \times 10^{-3}} \\ &= \mathbf{2.655 \times 10^{-8}\text{ F (or) } 26.55\text{ nF}} \end{aligned}$
(ii) $\begin{aligned}[t] C &= \frac{Q}{V} \\ Q &= C V \\ &= 2.655 \times 10^{-8} \times 120 \\ &= \mathbf{3.186 \times 10^{-6}\text{ C (or) } 3.186\text{ }\mu\text{C}} \end{aligned}$
(iii) $\begin{aligned}[t] V &= E d \\ E &= \frac{V}{d} \\ &= \frac{120}{1 \times 10^{-3}} = 120 \times 10^3 \\ &= \mathbf{1.2 \times 10^5\text{ V m}^{-1}} \end{aligned}$
Q: The plates of a parallel-plate capacitor are $50\text{ cm}^2$ in area and $1\text{ mm}$ apart, (i) What is its capacitance? (ii) When the capacitor is connected to a $24\text{ V}$ battery, what is the charge on each plate? (iii) What is the energy of the capacitor?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$A = 50\text{ cm}^2 = 50\,(10^{-2}\text{ m})^2 = 50 \times 10^{-4}\text{ m}^2$,
$d = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$, $V = 24\text{ V}$, $\kappa = 1$,
$\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}\text{ (or) F m}^{-1}$
(i) $\begin{aligned}[t] C &= \frac{\kappa \varepsilon_0 A}{d} \\ &= \frac{1 \times 8.85 \times 10^{-12} \times 50 \times 10^{-4}}{1 \times 10^{-3}} \\ &= \mathbf{44.25 \times 10^{-12}\text{ F (or) } 44.25\text{ pF}} \end{aligned}$
(ii) $\begin{aligned}[t] C &= \frac{Q}{V} \\ Q &= C V \\ &= 44.25 \times 10^{-12} \times 24 \\ &= \mathbf{1.062 \times 10^{-9}\text{ C (or) } 1.062\text{ nC}} \end{aligned}$
(iii) $\begin{aligned}[t] W &= \frac{1}{2} Q V \\ &= \frac{1}{2} \times 1.062 \times 10^{-9} \times 24 \\ &= \mathbf{1.2744 \times 10^{-8}\text{ J}} \end{aligned}$
Q: The capacitance of a parallel-plate capacitor is increased from $8\text{ }\mu\text{F}$ to $50\text{ }\mu\text{F}$ when a sheet of glass is inserted between its plates. What is the dielectric constant of the glass?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_0 = 8\text{ }\mu\text{F}$ (with vacuum or air),
$C = 50\text{ }\mu\text{F}$ (with glass plate), $\kappa = ?$
$\begin{aligned}[t] \kappa &= \frac{C}{C_0} \\ &= \frac{50\text{ }\mu\text{F}}{8\text{ }\mu\text{F}} = \mathbf{6.25} \end{aligned}$
Q: A parallel-plate capacitor of capacitance '$C$' is given the charge '$Q$' and then disconnected from the circuit. How much work is required to pull apart the plates of this capacitor to twice their original separation?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Original separation $= d$
Original capacitance $= C = \frac{\kappa \varepsilon_0 A}{d}$
New separation $= d' = 2d$
New capacitance $= C' = \frac{\kappa \varepsilon_0 A}{d'}$
$\begin{aligned}[t]
C' &= \frac{\kappa \varepsilon_0 A}{2d} \\
&= \frac{1}{2} \frac{\kappa \varepsilon_0 A}{d} \\
&= \mathbf{\frac{1}{2}C}
\end{aligned}$
Original energy stored $= W = \frac{1}{2}\frac{Q^2}{C}$
New energy stored $= W' = \frac{1}{2}\frac{Q^2}{C'}$
$\begin{aligned}[t] W' &= \frac{1}{2} \frac{Q^2}{\frac{1}{2}C} \\ &= \frac{Q^2}{C} \end{aligned}$
Required work $= W' - W$
$\begin{aligned}[t]
&= \frac{Q^2}{C} - \frac{1}{2}\frac{Q^2}{C} = \mathbf{\frac{1}{2}\frac{Q^2}{C}}
\end{aligned}$
Q: What potential difference must be applied across $10\text{ }\mu\text{F}$ capacitor if it is to have an energy content of $1\text{ J}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C = 10\text{ }\mu\text{F} = 10 \times 10^{-6}\text{ F}$, $W = 1\text{ J}$, $V = ?$
$\begin{aligned}[t] W &= \frac{1}{2} C V^2 \\ V^2 &= \frac{2W}{C} \\ V &= \sqrt{\frac{2W}{C}} = \sqrt{\frac{2 \times 1}{10 \times 10^{-6}}} = \mathbf{447.2\text{ V}} \end{aligned}$
Q: A charged capacitor of capacitance $C = 35\text{ }\mu\text{F}$ is discharged through a resistor of resistance $R = 120\text{ }\Omega$. (i) What is the time constant '$\tau$' of the discharging process? (ii) What is the elapsed time when the voltage falls to $10\text{ \%}$ of its original value?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C = 35\text{ }\mu\text{F} = 35 \times 10^{-6}\text{ F}$, $R = 120\text{ }\Omega$
(i) time constant, $\tau = ?$ $\quad$ (ii) $t = ?$
voltage falls to $10\text{ \%}$ of its original value,
$V_c = 10\text{ \% } V_0 = 0.1\text{ } V_0$
(i) $\begin{aligned}[t] \tau &= R C \\ &= 120 \times 35 \times 10^{-6} \\ &= \mathbf{4.2 \times 10^{-3}\text{ s (or) } 4.2\text{ ms}} \end{aligned}$
(ii) For discharging,
$\begin{aligned}[t]
V_c &= V_0 \left(e^{-t / R C}\right) \\
V_c &= V_0 \left(e^{-t / \tau}\right) \\
0.1\text{ } V_0 &= V_0 \left(e^{-t / \tau}\right) \\
0.1 &= e^{-t / \tau} \\
\ln 0.1 &= \ln e^{-t / \tau} \\
\ln 0.1 &= -\frac{t}{\tau} \\
t &= -\ln 0.1 \times \tau \\
&= -\ln 0.1 \times 4.2 \times 10^{-3} \\
&= \mathbf{9.67 \times 10^{-3}\text{ s (or) } 9.67\text{ ms}}
\end{aligned}$
Q: In a capacitor charging RC circuit, $C = 50\text{ }\mu\text{F}$. What is the value of resistance '$R$' that would produce a voltage **rise** to $20\text{ \%}$ of supply voltage after $1\text{ s}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C = 50\text{ }\mu\text{F} = 50 \times 10^{-6}\text{ F}$, $R = ?$, $t = 1\text{ s}$
voltage **rise** to $20\text{ \%}$ of supply voltage, $V_c = 20\text{ \% } V_0 = 0.2\text{ } V_0$
For charging,
$\begin{aligned}[t]
V_c &= V_0 \left(1 - e^{-t / R C}\right) \\
0.2\text{ } V_0 &= V_0 \left(1 - e^{-t / R C}\right) \\
0.2 &= 1 - e^{-t / R C} \\
e^{-t / R C} &= 1 - 0.2 = 0.8 \\
\ln \left(e^{-t / R C}\right) &= \ln 0.8 \\
-\frac{t}{R C} &= \ln 0.8 \\
-t &= \ln 0.8 \times R C \\
R &= \frac{-t}{\ln 0.8 \times C} \\
&= \frac{-1}{\ln 0.8 \times 50 \times 10^{-6}} \\
&= \mathbf{89.628 \times 10^3\text{ }\Omega = 89.628\text{ k}\Omega}
\end{aligned}$
Q: Three capacitors have capacitances of $5\text{ }\mu\text{F}$, $10\text{ }\mu\text{F}$ and $15\text{ }\mu\text{F}$.
(i) Find the equivalent capacitance when they are connected in parallel.
(ii) Find the equivalent capacitance when they are connected in series.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_1 = 5\text{ }\mu\text{F}$, $C_2 = 10\text{ }\mu\text{F}$, $C_3 = 15\text{ }\mu\text{F}$
(i) When $C_1$, $C_2$ and $C_3$ are connected in parallel
$\begin{aligned}[t]
C &= C_1 + C_2 + C_3 \\
&= 5 + 10 + 15 = \mathbf{30\text{ }\mu\text{F}}
\end{aligned}$
(ii) When $C_1$, $C_2$ and $C_3$ are connected in series
$\begin{aligned}[t]
\frac{1}{C} &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \\
&= \frac{1}{5} + \frac{1}{10} + \frac{1}{15} \\
&= \frac{6 + 3 + 2}{30} = \frac{11}{30} \\
C &= \frac{30}{11} = \mathbf{2.727\text{ }\mu\text{F}}
\end{aligned}$
Q: Find the capacitance that can be obtained by combining **three** $10\text{ }\mu\text{F}$ capacitors in **all possible ways**.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_1 = C_2 = C_3 = 10\text{ }\mu\text{F}$
$\Rightarrow$ When $C_1$, $C_2$ and $C_3$ are connected in parallel
$\begin{aligned}[t]
C &= C_1 + C_2 + C_3 \\
&= 10 + 10 + 10 = \mathbf{30\text{ }\mu\text{F}}
\end{aligned}$
$\Rightarrow$ When $C_1$, $C_2$ and $C_3$ are connected in series
$\begin{aligned}[t]
\frac{1}{C} &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \\
&= \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10} \\
C &= \frac{10}{3} = \mathbf{3.33\text{ }\mu\text{F}}
\end{aligned}$
$\Rightarrow$ When $C_1$ and $C_2$ are connected in parallel ($C_p$) and connected in series with $C_3$
$\begin{aligned}[t]
C_p &= C_1 + C_2 = 10 + 10 = 20\text{ }\mu\text{F} \\
\frac{1}{C} &= \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{20} + \frac{1}{10} = \frac{3}{20} \\
C &= \frac{20}{3} = \mathbf{6.67\text{ }\mu\text{F}}
\end{aligned}$
$\Rightarrow$ When $C_1$ and $C_2$ are connected in series ($C_s$) and connected in parallel with $C_3$
$\begin{aligned}[t]
\frac{1}{C_s} &= \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \implies C_s = 5\text{ }\mu\text{F} \\
C &= C_s + C_3 = 5 + 10 = \mathbf{15\text{ }\mu\text{F}}
\end{aligned}$
Q: The equivalent capacitance is $10\text{ }\mu\text{F}$ when '$n$' identical capacitors are connected in parallel and $0.4\text{ }\mu\text{F}$ when they are connected in series. Determine '$n$' and the capacitance of each capacitor.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_p = 10\text{ }\mu\text{F}$, $C_s = 0.4\text{ }\mu\text{F}$
For $n$ identical capacitors,
$C_1 = C_2 = C_3 = \dots = C_n = C$
In parallel,
$\begin{aligned}[t]
C_p &= n\text{ }C \text{ }--------------\text{ (1)}
\end{aligned}$
In series,
$\begin{aligned}[t]
C_s &= \frac{C}{n} \text{ }--------------\text{ (2)}
\end{aligned}$
eq (1) $\div$ eq (2)
$\begin{aligned}[t]
\frac{C_p}{C_s} &= \frac{n\text{ }C}{\frac{C}{n}} \\
\frac{C_p}{C_s} &= n\text{ }C \times \frac{n}{C} \\
\frac{10}{0.4} &= n^2 \\
n &= \sqrt{\frac{10}{0.4}} = \mathbf{5\text{ capacitors}}
\end{aligned}$
$n = 5$ in eq (2)
$\begin{aligned}[t]
0.4 &= \frac{C}{5} \\
C &= 0.4 \times 5 \\
&= \mathbf{2\text{ }\mu\text{F}}
\end{aligned}$
Q: A $35\text{ }\mu\text{F}$ capacitor is needed, but only $10\text{ }\mu\text{F}$ capacitors are available. How should a minimum number of $10\text{ }\mu\text{F}$ capacitors be connected so that the combination has a capacitance of $35\text{ }\mu\text{F}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
To get $35\text{ }\mu\text{F}$
$C_1 = C_2 = C_3 = \dots = C_n = C = 10\text{ }\mu\text{F}$
When $C_1$ and $C_2$ are connected in series,
$\begin{aligned}[t]
\frac{1}{C_s} &= \frac{1}{C_1} + \frac{1}{C_2} \\
&= \frac{1}{10} + \frac{1}{10} \\
&= \frac{1 + 1}{10} = \frac{2}{10} \\
C_s &= \frac{10}{2} = \mathbf{5\text{ }\mu\text{F}}
\end{aligned}$
When $C_3$, $C_4$ and $C_5$ are connected in parallel,
$\begin{aligned}[t]
C_p &= C_3 + C_4 + C_5 \\
&= 10 + 10 + 10 = \mathbf{30\text{ }\mu\text{F}}
\end{aligned}$
When $C_p$ and $C_s$ are connected in parallel,
$\begin{aligned}[t]
C &= C_p + C_s \\
&= 30 + 5 = \mathbf{35\text{ }\mu\text{F}}
\end{aligned}$
To get $35\text{ }\mu\text{F}$, minimum number of $10\text{ }\mu\text{F}$ capacitors = $\mathbf{5}$
Q: There capacitors have capacitance of $3\text{ }\mu\text{F}$, $10\text{ }\mu\text{F}$ and $15\text{ }\mu\text{F}$. How should they be connected to obtain the equivalent capacitances of (i) $2\text{ }\mu\text{F}$ (ii) $9\text{ }\mu\text{F}$ (iii) $12.5\text{ }\mu\text{F}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_1 = 3\text{ }\mu\text{F}$, $C_2 = 10\text{ }\mu\text{F}$, $C_3 = 15\text{ }\mu\text{F}$
(i) To get $2\text{ }\mu\text{F}$,
$C_1$, $C_2$ and $C_3$ are connected in series
$\begin{aligned}[t]
\frac{1}{C} &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \\
&= \frac{1}{3} + \frac{1}{10} + \frac{1}{15} \\
&= \frac{10 + 3 + 2}{30} = \frac{15}{30} \\
C &= \frac{30}{15} = \mathbf{2\text{ }\mu\text{F}}
\end{aligned}$
(ii) To get $9\text{ }\mu\text{F}$,
$C_2$ and $C_3$ are connected in series
$\begin{aligned}[t]
\frac{1}{C_s} &= \frac{1}{C_2} + \frac{1}{C_3} \\
&= \frac{1}{10} + \frac{1}{15} \\
&= \frac{3 + 2}{30} = \frac{5}{30} \\
C_s &= \frac{30}{5} = 6\text{ }\mu\text{F}
\end{aligned}$
$C_s$ and $C_1$ are connected in parallel
$\begin{aligned}[t]
C &= C_s + C_1 = 6 + 3 = \mathbf{9\text{ }\mu\text{F}}
\end{aligned}$
(iii) To get $12.5\text{ }\mu\text{F}$,
$C_1$ and $C_3$ are connected in series
$\begin{aligned}[t]
\frac{1}{C_s} &= \frac{1}{C_1} + \frac{1}{C_3} \\
&= \frac{1}{3} + \frac{1}{15} \\
&= \frac{5 + 1}{15} = \frac{6}{15} \\
C_s &= \frac{15}{6} = 2.5\text{ }\mu\text{F}
\end{aligned}$
$C_s$ and $C_2$ are connected in parallel
$\begin{aligned}[t]
C &= C_s + C_2 = 2.5 + 10 = \mathbf{12.5\text{ }\mu\text{F}}
\end{aligned}$
Q: Three capacitors of capacitance $3\text{ }\mu\text{F}$, $10\text{ }\mu\text{F}$ and $15\text{ }\mu\text{F}$ are connected in series with $100\text{ V}$ battery. What is the charge and the potential difference on each capacitor?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_1 = 3\text{ }\mu\text{F}$, $C_2 = 10\text{ }\mu\text{F}$, $C_3 = 15\text{ }\mu\text{F}$, $V = 100\text{ V}$
$Q_1 = ?,\quad Q_2 = ?,\quad Q_3 = ?$
$V_1 = ?,\quad V_2 = ?,\quad V_3 = ?$
When $C_1$, $C_2$ and $C_3$ are connected in series,
$\begin{aligned}[t]
\frac{1}{C} &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \\
&= \frac{1}{3} + \frac{1}{10} + \frac{1}{15} \\
&= \frac{10 + 3 + 2}{30} = \frac{15}{30} \\
C &= \frac{30}{15} = \mathbf{2\text{ }\mu\text{F}}
\end{aligned}$
$\begin{aligned}[t] C &= \frac{Q}{V} \\ Q &= C\text{ }V \\ &= 2 \times 10^{-6} \times 100 \\ &= \mathbf{200 \times 10^{-6}\text{ C}} \end{aligned}$
$Q_1 = Q_2 = Q_3 = Q = \mathbf{200 \times 10^{-6}\text{ C}}$ (in series)
$\begin{aligned}[t] V_1 &= \frac{Q_1}{C_1} = \frac{200 \times 10^{-6}}{3 \times 10^{-6}} = \mathbf{66.67\text{ V}} \\ V_2 &= \frac{Q_2}{C_2} = \frac{200 \times 10^{-6}}{10 \times 10^{-6}} = \mathbf{20\text{ V}} \\ V_3 &= \frac{Q_3}{C_3} = \frac{200 \times 10^{-6}}{15 \times 10^{-6}} = \mathbf{13.33\text{ V}} \end{aligned}$
Q: A capacitor having a capacitance of $2\text{ }\mu\text{F}$ and a charge of $2000\text{ }\mu\text{C}$ is connected in series with another capacitor having a capacitance of $8\text{ }\mu\text{F}$ and a charge of $1600\text{ }\mu\text{C}$.
(i) Find the potential difference of the individual capacitors prior to the connection.
(ii) Find the potential difference of the individual capacitors after the connection.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
$C_1 = 2\text{ }\mu\text{F},\quad C_2 = 8\text{ }\mu\text{F}$
$Q_1 = 2000\text{ }\mu\text{C},\quad Q_2 = 1600\text{ }\mu\text{C}$
(i) $V_1 = ?,\quad V_2 = ?$ (prior to the connection)
(ii) $V'_1 = ?,\quad V'_2 = ?$ (after the connection)
(i) Prior to the connection
$\begin{aligned}[t]
V_1 &= \frac{Q_1}{C_1} = \frac{2000 \times 10^{-6}}{2 \times 10^{-6}} = \mathbf{1000\text{ V}} \\
V_2 &= \frac{Q_2}{C_2} = \frac{1600 \times 10^{-6}}{8 \times 10^{-6}} = \mathbf{200\text{ V}}
\end{aligned}$
(ii) After the connection
$\begin{aligned}[t]
V &= V_1 + V_2 = 1000 + 200 = 1200\text{ V}
\end{aligned}$
When $C_1$ and $C_2$ are connected in series,
$\begin{aligned}[t]
\frac{1}{C} &= \frac{1}{C_1} + \frac{1}{C_2} \\
&= \frac{1}{2} + \frac{1}{8} \\
&= \frac{4 + 1}{8} = \frac{5}{8} \\
C &= \frac{8}{5} = 1.6\text{ }\mu\text{F}
\end{aligned}$
$\begin{aligned}[t] Q &= C\text{ }V \\ &= 1.6 \times 10^{-6} \times 1200 \\ &= 1920 \times 10^{-6}\text{ C} \end{aligned}$
$Q'_1 = Q'_2 = Q = 1920 \times 10^{-6}\text{ C}$ (in series)
$\begin{aligned}[t] V'_1 &= \frac{Q'_1}{C_1} = \frac{1920 \times 10^{-6}}{2 \times 10^{-6}} = \mathbf{960\text{ V}} \\ V'_2 &= \frac{Q'_2}{C_2} = \frac{1920 \times 10^{-6}}{8 \times 10^{-6}} = \mathbf{240\text{ V}} \end{aligned}$