Respuesta :
Given data:
* The current through the system is I = 3 A.
* The resistance of resistors connected in parallel is,
[tex]\begin{gathered} R_1=3\text{ ohm} \\ R_2=15\text{ ohm} \end{gathered}[/tex]Solution:
The equivalent resistance of the system is,
[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]Substituting the known values,
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{3}+\frac{1}{15} \\ \frac{1}{R_{eq}}=\frac{5+1}{15} \\ \frac{1}{R_{eq}}=\frac{6}{15} \end{gathered}[/tex]By taking inverse value,
[tex]\begin{gathered} R_{eq}=\frac{15}{4} \\ R_{eq}=3.75\text{ ohm} \end{gathered}[/tex]According to Ohm's law, the voltage across the battery is,
[tex]V=IR_{eq}[/tex]Substituting the known values,
[tex]\begin{gathered} V=3\times3.75 \\ V=11.25\text{ volts} \end{gathered}[/tex]Thus, the voltage across the battery is 11.25 volts.
