Answer :
We start by recalling Ohm's law, which states that
[tex]$$
V = I \cdot R.
$$[/tex]
Here, [tex]$V$[/tex] is the voltage, [tex]$I$[/tex] is the current, and [tex]$R$[/tex] is the resistance. To find the resistance, we solve for [tex]$R$[/tex] by rearranging the equation:
[tex]$$
R = \frac{V}{I}.
$$[/tex]
Given that the battery voltage is [tex]$6$[/tex] volts and that we want a current of [tex]$120$[/tex] amps, we substitute the values into the equation:
[tex]$$
R = \frac{6}{120}.
$$[/tex]
Performing the division, we find:
[tex]$$
R = 0.05 \text{ ohm}.
$$[/tex]
Thus, the resistance that would produce a current of [tex]$120$[/tex] amps from a [tex]$6$[/tex] V battery is [tex]$0.05$[/tex] ohm.
[tex]$$
V = I \cdot R.
$$[/tex]
Here, [tex]$V$[/tex] is the voltage, [tex]$I$[/tex] is the current, and [tex]$R$[/tex] is the resistance. To find the resistance, we solve for [tex]$R$[/tex] by rearranging the equation:
[tex]$$
R = \frac{V}{I}.
$$[/tex]
Given that the battery voltage is [tex]$6$[/tex] volts and that we want a current of [tex]$120$[/tex] amps, we substitute the values into the equation:
[tex]$$
R = \frac{6}{120}.
$$[/tex]
Performing the division, we find:
[tex]$$
R = 0.05 \text{ ohm}.
$$[/tex]
Thus, the resistance that would produce a current of [tex]$120$[/tex] amps from a [tex]$6$[/tex] V battery is [tex]$0.05$[/tex] ohm.
