Answer:
The maximum voltage that can be applied without damaging the resistor is 4.85 V
Explanation:
Electric Power in a Resistor
Given a resistor or resistance R connected to a circuit of voltage V carrying a current I. The relation between these three magnitudes is given by Ohm's Law:
V = R.I
The dissipated power P of a resistor can be calculated by the following equation, known as Joule's first law:
[tex]P = I^2.R[/tex]
Solving the first equation for I:
[tex]\displaystyle I=\frac{V}{R}[/tex]
Substituting in the second equation:
[tex]\displaystyle P=\frac{V^2}{R^2}.R[/tex]
Simplifying:
[tex]\displaystyle P=\frac{V^2}{R}[/tex]
Solving for V:
[tex]V=\sqrt{P.R}[/tex]
The resistor has a resistance of R=47Ω and can hold a maximum power of P=0.5 W, thus the maximum voltage is:
[tex]V=\sqrt{0.5\cdot 47}[/tex]
[tex]V=\sqrt{23.5}[/tex]
V = 4.85 V
The maximum voltage that can be applied without damaging the resistor is 4.85 V