Solveeit Logo

Question

Question: A \(1\mu F\) and a \(2\mu F\) capacitor are connected in series across a \(1200\,{\rm{V}}\) supply. ...

A 1μF1\mu F and a 2μF2\mu F capacitor are connected in series across a 1200V1200\,{\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 1400/3μC1400\,/3\,\mu C and 3200/3μC3200\,/3\,\mu C, and the potential difference across each capacitor is 1600/3V1600\,/3\,{\rm{V}}
B. Charges on capacitors are 1600/3μC1600\,/3\,\mu C and 3200/3μC3200\,/3\,\mu C, and the potential difference across each capacitor is 1600/3V1600\,/3\,{\rm{V}}
C. Charge on each capacitor is 1600μC1600\,\mu C and potential difference across each capacitor is 800V800\,{\rm{V}}.
D. Charge and potential difference across each capacitor are zero

Explanation

Solution

In this solution, find out the charge from those charged capacitors which are 1μF1\mu F and a 2μF2\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 1μF1\mu F and a 2μF2\mu F
Voltage, 1200V1200\,{\rm{V}}
Step 1:
1200q1q2=0\1200=q1+q2\1200=32q\q=23×1200\q=800μC\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 800μC800\,\mu C
Step 2: Finding V1{{\rm{V}}_{\rm{1}}} and V2{{\rm{V}}_2}
V1=qC1=8001=800V{V_1} = \dfrac{q}{{{C_1}}} = \dfrac{{800}}{1} = 800\,V
V2=qC2=8002=400V{V_2} = \dfrac{q}{{{C_2}}} = \dfrac{{800}}{2} = 400\,V
Step 3:
Here,
q1+q2=1600{q'_1} + {q'_2} = 1600 …… (i)
So, q11q22=0\dfrac{{{{q'}_1}}}{1} - \dfrac{{{{q'}_2}}}{2} = 0
Or,
q2=2q1{q'_2} = 2{q'_1}
From equation (i),
3q1=1600q1=16003μC\3q2=32003μC\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,
V=q1+q2C1+C2 =16003V\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 1600/3μC1600\,/3\,\mu C and 3200/3μC3200\,/3\,\mu C, and the potential difference across each capacitor is 1600/3V1600\,/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.