Question

Find the potential difference $V_a-V_b$ between the points a and b shown in each part of the figure (31-E14).

Answer 1

(a) $q=q_1+q_2$ ...(i)



On applying Kirchhoff's voltage law in the loop CabDC, we get

$$\begin{gathered} \frac{q_2}{2}+\frac{q_2}{4}-\frac{q_1}{4}=0 \\ \Rightarrow 2q_2+q_2-q_1=0 \\ \Rightarrow 3q_2=q_1...\left(\mathrm{ii}\right) \end{gathered}$$

On applying Kirchhoff's voltage law in the loop DCBAD, we get

$\frac{q}{2}+\frac{q_1}{4}-12=0$


$$\begin{gathered} \Rightarrow \frac{q_1+q_2}{2}+\frac{q_1}{4}-12=0 \\ \Rightarrow 3q_1+2q_2=48...\left(\mathrm{iii}\right) \end{gathered}$$

From eqs. (ii) and (iii), we get

$$\begin{gathered} 9q_2+2q_2=48 \\ \Rightarrow 11q_2=48 \\ \Rightarrow q_2=\frac{48}{11} \\ \mathrm{Now}, \\ {\mathrm{V}}_a-{\mathrm{V}}_b=\frac{q_2}{4\mathrm{\mu F}}=\frac{48}{44}=\frac{12}{11}\mathrm{V} \end{gathered}$$

(b) Let the charge in the loop be q.

Now, on applying Kirchhoff's voltage law in the loop, we get

$\frac{q}{2}+\frac{q}{4}-24+12=0$


$$\begin{gathered} \Rightarrow \frac{3q}{4}=12 \\ \Rightarrow q=16\mathrm{\mu C} \\ \mathrm{Now}, \\ {V}_a-V_b=\frac{-q}{2\mathrm{\mu F}} \\ \Rightarrow V_a-V_b=\frac{-16\mathrm{\mu C}}{2\mathrm{\mu F}}=-8\mathrm{V} \end{gathered}$$

(c) $V_{\mathrm{a}}-V_{\mathrm{b}}=2-\frac{2-q}{2\mathrm{\mu F}}$


In the loop,

$$\begin{gathered} 2+2-\frac{q}{2}-\frac{q}{2}=0 \\ \Rightarrow q=4\mathrm{C} \end{gathered}$$

$\therefore V_{\mathrm{a}}-V_{\mathrm{b}}=2-\frac{4}{2}=2-2=0\mathrm{V}$

(d)



Net charge flowing through all branches, q = 24 + 24 + 24 = 72 μC
Net capacitance of all branches, C = 4 + 2 + 1 = 7 μF
The total potential difference (V) between points a and b is given by
$$\begin{gathered} V=\frac{q}{C} \\ \Rightarrow V=\frac{72}{7}=10.3\mathrm{V} \end{gathered}$$
As the negative terminals of the batteries are connected to a, the net potential between points a and b is $-$10.3 V.