Answer:

Explanation:

When the current through a resistor is increased, the power dissipated in the resistor will increase. The power dissipated in a resistor can be calculated using the formula:

\[P = I^2R\]

Where:

- \(P\) is the power in watts (W).

- \(I\) is the current in amperes (A).

- \(R\) is the resistance in ohms (Ω).

Now, let's consider the scenario where the current is increased by 200%. This means the new current (\(I_{\text{new}}\)) is 200% (or 2 times) the original current (\(I_{\text{original}}\)).

\(I_{\text{new}} = 2 \times I_{\text{original}}\)

Now, let's calculate the new power (\(P_{\text{new}}\)) with the increased current:

\[P_{\text{new}} = (I_{\text{new}})^2 \cdot R\]

Substitute \(I_{\text{new}}\) with \(2 \cdot I_{\text{original}}\):

\[P_{\text{new}} = (2 \cdot I_{\text{original}})^2 \cdot R\]

Simplify:

\[P_{\text{new}} = 4 \cdot (I_{\text{original}}^2) \cdot R\]

Now, let's compare the new power (\(P_{\text{new}}\)) to the original power (\(P_{\text{original}}\)):

\[\text{Increase in power} = P_{\text{new}} - P_{\text{original}}\]

Substitute the expressions for \(P_{\text{new}}\) and \(P_{\text{original}}\):

\[\text{Increase in power} = (4 \cdot I_{\text{original}}^2 \cdot R) - (I_{\text{original}}^2 \cdot R)\]

Now, factor out \(I_{\text{original}}^2 \cdot R\):

\[\text{Increase in power} = I_{\text{original}}^2 \cdot R \cdot (4 - 1)\]

Simplify:

\[\text{Increase in power} = 3 \cdot I_{\text{original}}^2 \cdot R\]

So, when the current through the resistor is increased by 200%, the increase in power dissipated will be 3 times the original power dissipated.