Question

A $1\mu F$ and a $2\mu F$ capacitor are connected in series across a $1200\,{\rm{V}}$ supply. The charged capacitors are disconnected from the line and from each other and are now reconnected with the terminals of like sign together. Find the final charge on each capacitor and the voltage across each capacitor:
A. Charges on capacitors are $1400\,/3\,\mu C$ and $3200\,/3\,\mu C$, and the potential difference across each capacitor is $1600\,/3\,{\rm{V}}$
B. Charges on capacitors are $1600\,/3\,\mu C$ and $3200\,/3\,\mu C$, and the potential difference across each capacitor is $1600\,/3\,{\rm{V}}$
C. Charge on each capacitor is $1600\,\mu C$ and potential difference across each capacitor is $800\,{\rm{V}}$.
D. Charge and potential difference across each capacitor are zero

Answer 1

Hint:In this solution, find out the charge from those charged capacitors which are $1\mu F$ and a $2\mu F$. Then, find out the voltage from each capacitor and put all the value to find the voltage across the capacitor.

Complete step by step solution:
We have,
Capacitor $1\mu F$ and a $2\mu F$
Voltage, $1200\,{\rm{V}}$
Step 1:
$\begin{array}{c}1200 - \dfrac{q}{1} - \dfrac{q}{2} = 0\\1200 = \dfrac{q}{1} + \dfrac{q}{2}\\1200 = \dfrac{3}{2}q\\q = \dfrac{2}{3} \times 1200\\q = 800\,\mu C\end{array}$
Hence, the charge on each capacitor is $800\,\mu C$
Step 2: Finding ${{\rm{V}}_{\rm{1}}}$ and ${{\rm{V}}_2}$
${V_1} = \dfrac{q}{{{C_1}}} = \dfrac{{800}}{1} = 800\,V$
${V_2} = \dfrac{q}{{{C_2}}} = \dfrac{{800}}{2} = 400\,V$
Step 3:
Here,
${q'_1} + {q'_2} = 1600$ …… (i)
So, $\dfrac{{{{q'}_1}}}{1} - \dfrac{{{{q'}_2}}}{2} = 0$
Or,
${q'_2} = 2{q'_1}$
From equation (i),
$\begin{array}{c}3{{q'}_1} = 1600\\{{q'}_1} = \dfrac{{1600}}{3}\,\mu C\\3{{q'}_2} = \dfrac{{3200}}{3}\,\mu C\end{array}$
Hence,
$\begin{array}{c}V = \dfrac{{{q_1} + {q_2}}}{{{C_1} + {C_2}}}\\ = \dfrac{{1600}}{3}\,V\end{array}$
So, the required answer is option B Charges on capacitors are $1600\,/3\,\mu C$ and $3200\,/3\,\mu C$, and the potential difference across each capacitor is $1600\,/3\,{\rm{V}}$.

Additional information:
Capacitor: A capacitor is an electrical energy storage device in an electric field. It has two terminals and is a passive electronic component. Capacitance is the consequence of a condenser. Initially, the condenser was called a condenser or condensation. A capacitor is a passive electrical, two-terminal component used to store electrostatic energy in an electric field. A condenser is not dissipating energy, unlike a resistance. A condenser instead stores energy between its plates as an electrostatic field.
Terminals: A terminal is the end of a driver from a part, computer or network. Electrons flow from the negative terminal to the positive terminal from a galvanic cell, like an ordinary AA battery, while the normal current is the opposite. Terminals are pins in a connector that supplies electricity to protect the connections. Almost always made up of metal, some use other conductive materials (carbon, silicon, etc.).


Note:Since the capacitance is in micro, so the amount of charge will also be in micro-units.