Question

a 1.5 kω resistor is found to have 22.5 v across it. (a) what is the current in the resistor? ________ (b) what is the power dissipated in the resistor? ________ (c) could a 1/4 w resistor be used in this application? explain your answer

Explanation:

Step1: Use Ohm's law for current

Ohm's law is $I=\frac{V}{R}$. Given $V = 22.5\ V$ and $R=1.5\times10^{3}\ \Omega$, then $I=\frac{22.5}{1.5\times 10^{3}}$.
$I = 15\times10^{-3}\ A=15\ mA$

Step2: Use power - current - voltage relation for power

The power formula is $P = VI$. Substitute $V = 22.5\ V$ and $I = 15\times10^{-3}\ A$. So $P=22.5\times15\times 10^{-3}$.
$P = 0.3375\ W$

Step3: Compare power ratings

The power dissipated in the resistor is $P = 0.3375\ W$. A $\frac{1}{4}\ W=0.25\ W$ resistor is considered. Since $0.3375\ W>0.25\ W$, a $\frac{1}{4}\ W$ resistor cannot be used as it will over - heat.

Answer:

(a) $15\ mA$
(b) $0.3375\ W$
(c) No, because the power dissipated in the resistor ($0.3375\ W$) is greater than the power rating of a $\frac{1}{4}\ W$ resistor.