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Q.
In the given circuit below, the points $A, B$ and $C$ are at same potential. If the potential difference between $B$ and $D$ is $30\, V$, then the potential difference between $A$ and $O$ is
Potential difference between the points $O$ and $D$ $\left(V_{O D}\right)$ is 3 times the potential difference between $B$ and $O\left(V_{B O}\right)$.
From the figure,
$V_{B D}=V_{B O}+V_{O D}$ [Given]
$\Rightarrow V_{B D} =V_{B O}+3 V_{B O} \left[V_{O D}=3 V_{B O}\right]$
$\Rightarrow V_{B D} =4 V_{B O}$
$=V_{B O}=\frac{30}{4}=7.5\, V$
Now, points $A$ and $D$ have same potential.
So, $V_{A O}=V_{O D}=V_{B O}=7.5\, V$