To solve this problem, we need to use Ohm's law, which is given by the formula:
[tex]\[ V = IR \][/tex]
Where:
- [tex]\( V \)[/tex] is the voltage,
- [tex]\( I \)[/tex] is the current,
- [tex]\( R \)[/tex] is the resistance.
Now, we are given the initial current [tex]\( I = 1.2 \)[/tex] A. According to the problem, the voltage decreases to one third of its original value while the resistance remains constant.
Let's denote the original voltage by [tex]\( V \)[/tex]. If the voltage decreases to one third of its original value, the new voltage [tex]\( V_{\text{new}} \)[/tex] will be:
[tex]\[ V_{\text{new}} = \frac{V}{3} \][/tex]
Since the resistance [tex]\( R \)[/tex] remains constant, we use Ohm's law to find the new current, [tex]\( I_{\text{new}} \)[/tex], corresponding to the new voltage:
[tex]\[ V_{\text{new}} = I_{\text{new}} R \][/tex]
Substituting [tex]\( V_{\text{new}} = \frac{V}{3} \)[/tex] into the equation, we get:
[tex]\[ \frac{V}{3} = I_{\text{new}} R \][/tex]
We also know from the original condition that:
[tex]\[ V = I R \][/tex]
Now, substituting [tex]\( V = I R \)[/tex] into the new voltage equation:
[tex]\[ \frac{I R}{3} = I_{\text{new}} R \][/tex]
We can simplify by canceling [tex]\( R \)[/tex] from both sides:
[tex]\[ \frac{I}{3} = I_{\text{new}} \][/tex]
Given that the initial current [tex]\( I \)[/tex] is [tex]\( 1.2 \)[/tex] A, the new current [tex]\( I_{\text{new}} \)[/tex] will be:
[tex]\[ I_{\text{new}} = \frac{1.2}{3} \][/tex]
So,
[tex]\[ I_{\text{new}} = 0.4 \][/tex] A
Therefore, if the voltage decreases to one third of its original amount, the resulting current will be:
[tex]\[ \boxed{0.4 \, \text{A}} \][/tex]
