Calculate the charges which will flow in sections $1$ and $2$ in Fig., when key $K$ is pressed.

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Solution

Initially when key K is open, the lower cell is not making any difference.
Hence we can ignore that part.
$C_{2}$ and $C_{1}$ are in series combination.
We know that whenever 2 or more capacitors are in series, then total capacitance is given by $\frac{1}{C} = {\frac{1}{C}}_{1} + {\frac{1}{C}}_{2}$
$∴ C = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$
$E_{1} C_{1} = E_{2} C_{2} = Q_{2}$ and $E = E_{1} + E_{2}$
Whenever a capacitor is connected to a D.C. source such as a cell, current never flows through the circuit. This is because a capacitor applies infinite impedance to DC.
However the capacitors get charged and remain as it is.
We know that Charge Q accumulated in a capacitor is given by $Q = C E$
$Q_{2} = \frac{E \times C_{1} C_{2}}{C_{1} + C_{2}}$
Now when the Key is pressed, the $C_{1}$ will get an additional charge of $E C_{1}$ .
Hence the total charge on $C_{1}$ is $E C_{1} + \frac{E \times C_{1} C_{2}}{C_{1} + C_{2}} = E C_{1} (1 + \frac{C_{2}}{C_{1} + C_{2}})$

Answer:- Charge flowing through section 2 is $Q_{2} = \frac{E \times C_{1} C_{2}}{C_{1} + C_{2}}$
Charge flowing through section 1 is $E C_{1} (1 + \frac{C_{2}}{C_{1} + C_{2}})$

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