What is the voltage drop across a 3 kω resistor connected to a 9v power source?

1 answer:

8 0

We connect a 3 $k\Omega$ resistor in series to a power source. With the series connection one end of the resistor is connected to the positive terminal of the power source while the other the other end of the resistor is connected to the negative terminal of the power source. Hence the current through the resistor is given bythe Ohm's law
$I=U/R$ and the voltage drop on the resistor is given equal to the power source voltage.
$U=9 V$

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Answer:

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Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

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$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$