Question

a resistor, of resistance 47 , is damaged when it dissipates a power greater than 0.5W. determine the maximum voltage that can be applied without damaging the resistor​

Answer 1

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:

$P = I^2.R$

Solving the first equation for I:

$\displaystyle I=\frac{V}{R}$

Substituting in the second equation:

$\displaystyle P=\frac{V^2}{R^2}.R$

Simplifying:

$\displaystyle P=\frac{V^2}{R}$

Solving for V:

$V=\sqrt{P.R}$

The resistor has a resistance of R=47Ω and can hold a maximum power of P=0.5 W, thus the maximum voltage is:

$V=\sqrt{0.5\cdot 47}$

$V=\sqrt{23.5}$

V = 4.85 V

The maximum voltage that can be applied without damaging the resistor is 4.85 V