Respuesta :
Given data:
* The total current in the system is I = 2.8 A.
* The resistance of resistors connected in parallel is,
[tex]\begin{gathered} R_1=14\text{ ohm} \\ R_2=17\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}{14}+\frac{1}{17} \\ \frac{1}{R_{eq}}=\frac{17+14}{14\times17} \\ R_{eq}=\frac{14\times17}{17+14} \\ R_{eq}=7.68\text{ ohm} \end{gathered}[/tex]According to Ohm's law, the voltage across the battery in terms of the current and equivalent resistance is,
[tex]\begin{gathered} V=IR_{eq} \\ V=2.8\times7.68 \\ V=21.5\text{ volts} \end{gathered}[/tex]Thus, the voltage across the battery is 21.5 volts.
