Given data:
* The current through the circuit is,
![I=4.4\text{ A}](https://img.qammunity.org/2023/formulas/physics/high-school/i7dh93exx3usr8iascxtm0t0e6y6nkjrc7.png)
* The value of resistances given is,
![\begin{gathered} R_1=10\text{ ohm} \\ R_2=11\text{ ohm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/zwnbv3vpvw09hjlf5572qeu2bg2o9whqau.png)
Solution:
The equivalent resistance of the resistors connected in parallel is,
![\begin{gathered} (1)/(R_(eq))=(1)/(R_1)+(1)/(R_2) \\ (1)/(R_(eq))=(1)/(10)+(1)/(11) \\ (1)/(R_(eq))=(11+10)/(110) \\ (1)/(R_(eq))=(21)/(110) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/vkauef8k2ceflu63sn18oecxdq4ryvr7po.png)
By simplifying,
![\begin{gathered} R_(eq)=(110)/(21) \\ R_(eq)=5.24\text{ ohm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/z70vjzko5b6ankixw1pagj3inwyclesup0.png)
According to Ohm's law, the voltage across the battery in terms of the current and equivalent resistance is,
![\begin{gathered} V=IR_(eq) \\ V=4.4*5.24 \\ V=23.06\text{ volts} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/uwkgo0cfq3q674yeew4ewt8osldiqucm8l.png)
Thus, the voltage across the battery is 23.06 volts.