Question 1.SGPYQ.5: In the figure given below, the potential difference between ......

At node P,

\begin{aligned} & 2+\frac{V_{ P }-10}{2}+\frac{V_{ P }}{8}=0 \\ & 16+4 V_{ P }-40+V_{ P }=0 \\ & 5 V_{ P }-24=0 \Rightarrow V_{ P }=\frac{24}{5} \end{aligned}

At node Q,

\begin{aligned} & 2=\frac{V_{ Q }-10}{4}+\frac{V_{ Q }-0}{6} \\ & 2=\frac{6 V_{ Q }-60+4 V_{ Q }}{24} \end{aligned}

48=10 V_{ Q }-60 \Rightarrow 24=5 V_{ Q }-30 \Rightarrow V_{ Q }=\frac{54}{5}

Therefore, the potential difference between P and Q is V_{ P }-V_{ Q }

V_{ P }-V_{ Q }=\frac{24}{5}-\frac{54}{5}=-6  V