If 80. joules of electrical energy is dissipated by a 10.-ohm resistor in 2.0 seconds, the current in theresistor is
Question
If 80. joules of electrical energy is dissipated by a 10.-ohm resistor in 2.0 seconds, the current in the resistor is
Solution
To find the current, we can use the formula for power, which is P = IV, where P is power (in watts), I is current (in amperes), and V is voltage (in volts).
We know that power can also be calculated by the formula P = E/t, where E is energy (in joules) and t is time (in seconds).
Given that E = 80 joules and t = 2.0 seconds, we can find P = 80J / 2.0s = 40 watts.
Now we can use Ohm's law, V = IR, where V is voltage, I is current, and R is resistance. We know that R = 10 ohms.
We can rearrange Ohm's law to solve for I: I = V/R.
But we don't have V. However, we know that power P = IV, so we can solve this equation for V: V = P/I.
Substituting P = 40 watts into this equation gives V = 40W / I.
Now we can substitute V = 40W / I into the rearranged Ohm's law equation: I = (40W / I) / 10 ohms.
Solving this equation for I gives I = sqrt(40W / 10 ohms) = 2 amperes.
So, the current in the resistor is 2 amperes.
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