Question

Find the potential difference between the points A and B as shown in the figure. (initially, all the capacitors are uncharged)

Answer 1

We need to find the potential difference between point A and point B. Between these two points in the branch AB, we have one capacitor and one emf source. It is given that initially, all the capacitors are uncharged so, at time t=0, the charge on any capacitor is zero.

Complete step by step answer: To solve this problem we assume let us grounded point A, so the potential at point A becomes zero, so we can represent the circuit as

Now we can apply charge conservation at point B, which says the total charge coming out at B must be equal to zero.
Let us assume charge emanating from point B in three branches up, middle and down respectively be ${{q}_{\begin{smallmatrix} 1 \\\\\\\\\end{smallmatrix}}}$,${{q}_{2}}$ and ${{q}_{3}}$
So by charge conservation ${{q}_{1}}+{{q}_{2}}+{{q}_{3}}=0$
Also, charge on any capacitor is given by Q=CV
So. Bu using this formula we find out the three charges as:
${{q}_{1}}=6\times {{10}^{-6}}({{V}_{b}}-10) \\\ {{q}_{2}}=4\times {{10}^{-6}}({{V}_{b}}-20) \\\ {{q}_{3}}=10\times {{10}^{-6}}({{V}_{b}}-30) \\\$
Adding them all we get, $0=6\times {{10}^{-6}}({{V}_{b}}-10)+4\times {{10}^{-6}}({{V}_{b}}-20)+10\times {{10}^{-6}}({{V}_{b}}-30)$
Solving this we find
${{V}_{b}}=\dfrac{440}{20}V \\\ =22V \\\$
So, the value of the potential value at point B is 22 V and we had assumed the value of the potential at point A to be zero, so the value of the potential difference between A and B comes out to be 22 V.

Note:
Potential is due to electric charge and we always measure the potential difference, while doing the calculation here we had assumed that potential at one point to be zero would not affect the measurement because we always measure the potential difference and not the potential.