The current will drop to half of its original value.
To solve this problem, we must use Ohm's law that states that in a circuit the voltage [tex]v[/tex] across a resistor is directly proportional to the current that flows through that circuit. In other words:
[tex]v \propto Ri \\ \\ or: \\ \\ v=Ri \\ \\ It's \ also \ valid: i=\frac{v}{R}[/tex]
According to our problem, if the voltage in the circuit is held constant while the resistance double, that is [tex]R_{New}=2R \ and \ v_{New}=v[/tex]. So:
[tex]v=R_{New}i_{New} \\ \\ v=2Ri_{New} \\ \\ Isolating \ i_{New}: \\ \\ i_{New}=\frac{v}{2R} \therefore i_{New}=\frac{1}{2}\frac{v}{R} \therefore i_{New}=\frac{1}{2}i[/tex]
In conclusion, the current will drop to half of its original value.
