A capacitor of capacity C is charged fully by a cell of emf V/2 and then it is
disconnected and again connected with a cell of emf V (+ve plate of capacitor with +ve terminal of cell and vice versa). The heat developed in connecting wire during charging by second cell is ??
Lulu
To find the heat developed in the connecting wire during charging by the second cell, we need to calculate the amount of charge transferred from the second cell to the capacitor.
First, let's consider the charging process with the first cell of emf V/2.
When the capacitor is fully charged by the first cell, it gains a charge Q equal to its capacity C multiplied by the voltage V/2, i.e., Q = C * (V/2) = CV/2.
The energy stored in the capacitor after being charged by the first cell is given by E = (1/2) * C * (V/2)^2 = (1/2) * C * V^2 / 4 = CV^2 / 8.
Now, let's move on to the second charging process with the cell of emf V.
When the capacitor is connected to the second cell, charge flows from the positive terminal of the cell to the positive plate of the capacitor, making the capacitor potential reach the emf of the cell (V).
Hence, the charge transferred from the second cell is Q' = C * V.
Now, we can calculate the heat developed in the connecting wire during charging by the second cell.
The heat generated in the wire is equal to the energy lost by the capacitor, which is given by the formula: Heat = E - E'.
Substituting the values, Heat = (CV^2 / 8) - (CV^2 / 2) = CV^2 / 8 - 4CV^2 / 8 = -3CV^2 / 8.
So, the heat developed in the connecting wire during charging by the second cell is -3CV^2 / 8.