Current I is passing through a load of resistance R. If the voltage across the load doubles and the resistance of the load is reduced by half then what is the amount of current passing through the load?
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Detailed Explanation
According to Ohm's law, the relationship between voltage, current and resistance is V = IR which can be rearranged as \(I = \frac{\mathrm{V} }{\mathrm{R} } \).
If the voltage doubles and the resistance is reduced by half then \(I_1 = \frac{\mathrm{2V} }{\mathrm{0.5 \Omega } } = 4 \frac{\mathrm{V} }{\mathrm{R} } = 4I\) .
Therefore, the current flowing through the load will increase four times.
If the voltage doubles and the resistance is reduced by half then \(I_1 = \frac{\mathrm{2V} }{\mathrm{0.5 \Omega } } = 4 \frac{\mathrm{V} }{\mathrm{R} } = 4I\) .
Therefore, the current flowing through the load will increase four times.
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