Answer:
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf Circuit & \bf Current \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 5 \\ \\ \sf 2 & \sf 1 \\ \\ \sf 3 & \sf 0.12 \\ \\ \sf 4 & \sf 1.5 \times {10}^{ - 2} \end{array}} \\ \end{gathered}[/tex]
Explanation:
According to ohm's law, potential difference is directly proportional to the applied current given by formula,
[tex]V = IR[/tex]
Where R is resistance.
can also be written as,
[tex]I = \frac{V}{R} [/tex]
Circuit 1,
[tex]I_1 = \frac{10}{2} = 5 A[/tex]
Circuit 2,
[tex]I_2 = \frac{12}{12} = 1A[/tex]
Circuit 3
[tex]I_3 = \frac{6}{50} = 0.12A[/tex]
Circuit 4,
[tex]I_4 = \frac{1.5}{100} = 1.5 \times {10}^{ - 2} A[/tex]
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