Answer :
First, recognize that in a series circuit the total resistance is the sum of all individual resistances. The circuit has three resistors of [tex]\(2\,\Omega\)[/tex] each and one resistor of [tex]\(6\,\Omega\)[/tex]. So, the total resistance [tex]\( R_{\text{total}} \)[/tex] is:
[tex]$$
R_{\text{total}} = 2\,\Omega + 2\,\Omega + 2\,\Omega + 6\,\Omega = 12\,\Omega.
$$[/tex]
Since the current [tex]\( I \)[/tex] in the circuit is given as [tex]\(4.0\,\text{A}\)[/tex], we use Ohm's law, which states:
[tex]$$
V = I \times R.
$$[/tex]
Substituting the known values:
[tex]$$
V = 4.0\,\text{A} \times 12\,\Omega = 48\,\text{V}.
$$[/tex]
Thus, the voltage of the battery is:
[tex]$$
\boxed{48\,\text{V}}.
$$[/tex]
[tex]$$
R_{\text{total}} = 2\,\Omega + 2\,\Omega + 2\,\Omega + 6\,\Omega = 12\,\Omega.
$$[/tex]
Since the current [tex]\( I \)[/tex] in the circuit is given as [tex]\(4.0\,\text{A}\)[/tex], we use Ohm's law, which states:
[tex]$$
V = I \times R.
$$[/tex]
Substituting the known values:
[tex]$$
V = 4.0\,\text{A} \times 12\,\Omega = 48\,\text{V}.
$$[/tex]
Thus, the voltage of the battery is:
[tex]$$
\boxed{48\,\text{V}}.
$$[/tex]
