If a conductor with a current of 200 mA passing through it converts 40 J of electrical energy into heat in 30 s, what is the potential drop across the conductor?
We use the relationship "P=\\frac{\\Delta E}{\\Delta t}= I \\Delta V" to find the potential drop ∆V:
"\\Delta V= \\frac{\\Delta E}{I\\Delta t}= \\frac{40\\,J}{(0.200 A)(30\\,s)}= \\frac{20}{3}V \\approxeq 6.67\\,V"
In conclusion, the potential drop across the conductor is ∆V=6.67 V.
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