**Hint: The potential drop on parallel connection is same.**

**Step 1: Equivalent resistance through ADC,**

The equivalent resistance in the series connection is defined as,

$R=R_{1}+R_{2}+R_{3}$

$R=5Ω+5Ω+5Ω$

$R=15Ω$

**Step 2: Finding the current through ADC,**

As the potential drop in the parallel connection, therefore the potential drop across $15Ω$ is same.

According to Ohm's law,

$V=IR$

$I=RV $

$I=152 $

**Step 3: Finding the potential drop.**

Using the Ohm's law, the potential drop across AD is,

$V=IR$

$V=152 ×5$

$V=32 $