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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?

A 4I

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.


Jonus Wang

2 years ago

Helped me a ton


2 years ago

Learning so fast with this app


2 years ago

Good help for the real test. I went in and took the practice test and there are similar questions on here that’s on the practice which is good.

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