Tecnología y Electrónica, pregunta formulada por 0xakanesoiunx0, hace 16 horas

14. If you wish to increase the amount of current in a resistor from 120 mA to 160 mA by changing the 24 V source, what should the new voltage setting be?
can u put the operations?, PLS PLS PLS I NEED IT T_T

Respuestas a la pregunta

Contestado por belmontDubois

Respuesta:

32V

Explicación:

The current value of the resistor is:

$R=\frac{V}{I}=\frac{24V}{120mA}=200\Omega$

The variable that will change will be the voltage, the value of the resistor will remains the same, so we aply Ohm's law again:

$V=R\times I=(200\Omega)(160mA)=32V$

The new voltage to apply should be: $32V$

0xakanesoiunx0: pls, can you answer me anothers questions? (3)

0xakanesoiunx0: 19. Four amperes of current are measured through a 24 resistor connected across a
voltage source. How much voltage does the source produce?
20. Twelve volts are applied across a resistor. A current of 3 mA is measured. What is the
value of the resistor?

0xakanesoiunx0: 23. A 120 V lamp-dimming circuit is controlled by a rheostat and protected from excessive
current by a 3 A fuse. To what minimum resistance value can the rheostat be set without
blowing the fuse? Assume a lamp resistance of 20 ohms.

0xakanesoiunx0: i would be so gratefull

belmontDubois: 19. V = R * I = (24)(4A)= 96V

belmontDubois: 20. R = V/I = 12V/3mA = 4kOhms

belmontDubois: 23. Rt = 120V/3A = 40 Ohms

belmontDubois: However, Rt = Rlamp + Rrheost = 40Ohms

belmontDubois: so, Rrheost = Rt - Rlamp = 40 - 20 = 20Ohms

0xakanesoiunx0: THANKSSSSSS

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