Question

The maximum current rating for a 10 kilo ohm 0.5w resistor is

Answer 1

$\huge\mathsf{Solution}$
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$As \: we \: know \: that \: the \: expression \: of \: Power \: P = {I}^{2}R$

$Now \: \ \: \\ \\ {I}^{2} = \frac{P}{R}$

${I}^{2} = \frac{0.5}{1000} = 5.0 \times {10}^{ - 5}$

$I \: = \sqrt{5 \times {10}^{ - 5} } \\ \\ = 0.00707 \\ \\ = 7.07 \times {10}^{ - 3}$

Resistor= 7.07 mA

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Answer 2

Explanation:

Given that,

Resistance of the resistor, $R=10\ k\Omega=10^4\ \Omega$

Power of the resistor, P = 0.5 watts

Let I is the maximum current flowing through the resistor. We know that the power of any resistor is given by :

$P=I^2R$

$I=\sqrt{\dfrac{P}{R}}$

$I=\sqrt{\dfrac{0.5}{10^4}}$

I = 0.00707 A

or

I = 7.07 mA

So, the maximum current rating for the resistor is 7.07 mA. Hence, this is the required solution.