Given data:
* The current through the circuit is,
[tex]I=4.4\text{ A}[/tex]* The value of resistances given is,
[tex]\begin{gathered} R_1=10\text{ ohm} \\ R_2=11\text{ ohm} \end{gathered}[/tex]Solution:
The equivalent resistance of the resistors connected in parallel is,
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{11} \\ \frac{1}{R_{eq}}=\frac{11+10}{110} \\ \frac{1}{R_{eq}}=\frac{21}{110} \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} R_{eq}=\frac{110}{21} \\ R_{eq}=5.24\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=4.4\times5.24 \\ V=23.06\text{ volts} \end{gathered}[/tex]Thus, the voltage across the battery is 23.06 volts.
