Answer:
The original voltage in the circuit is [tex]\frac{81}{125}[/tex]
Step-by-step explanation:
We are given
Current voltage is
[tex]V_c=\frac{27}{125}[/tex]
Let's assume original voltage as Vo
we are given
it has been reduced 3 times
so,
[tex]V_c=\frac{V_o}{3}[/tex]
we can plug values
[tex]\frac{27}{125}=\frac{V_o}{3}[/tex]
Multiply both sides by 3
[tex]\frac{27}{125}\times 3=\frac{V_o}{3}\times 3[/tex]
[tex]\frac{81}{125}=V_o[/tex]
[tex]V_o=\frac{81}{125}[/tex]
So,
The original voltage in the circuit is [tex]\frac{81}{125}[/tex]
