An industrial temperature sensor module with a calibrated internal resistance of 4.5\text{… — Armed Services Vocational Aptitude Battery (ASVAB)

An industrial temperature sensor module with a calibrated internal resistance of $4.5\text{ k}\Omega$ draws a steady direct current of $8\text{ mA}$ during normal operation. What is the voltage drop across the sensor module in volts ( $\text{V}$ )?

Cevap: 36 V

Cevap

The voltage drop across the sensor module is

36\text{ V}

According to Ohm's Law, voltage (

V

) equals current (

I

) multiplied by resistance (

R

). Given

I = 8\text{ mA} = 0.008\text{ A}

and

R = 4.5\text{ k}\Omega = 4500\text{ }\Omega

, the calculation is

V = 0.008\text{ A} \times 4500\text{ }\Omega = 36\text{ V}

Adım Adım Çözüm

Convert the given metric prefixes to standard base units (amperes and ohms).

I = 8\text{ mA} = 8 \times 10^{-3}\text{ A} = 0.008\text{ A}

and

R = 4.5\text{ k}\Omega = 4.5 \times 10^{3}\text{ }\Omega = 4500\text{ }\Omega

Ohm's law requires fundamental SI units (volts, amperes, and ohms) for accurate mathematical evaluation.

Select the appropriate form of Ohm's Law to calculate voltage (

V

V = I \times R

Voltage across a resistive load equals the product of the current flowing through it and its resistance.

Substitute the converted values into the formula and solve.

V = 0.008\text{ A} \times 4500\text{ }\Omega = 36\text{ V}

Multiplying

8 \times 10^{-3}\text{ A}

4.5 \times 10^{3}\text{ }\Omega

simplifies directly to

8 \times 4.5 = 36\text{ V}

because the metric prefixes milli (

10^{-3}

) and kilo (

10^{3}

) cancel each other.

Anahtar Kavram

Ohm's Law: Voltage Calculation (

V = I \times R

)

Alternatif Yöntem

Recognize that the prefix multiplier milli- (

10^{-3}

) and kilo- (

10^{3}

) cancel out when multiplied together:

(8\text{ mA}) \times (4.5\text{ k}\Omega) = 8 \times 4.5\text{ V} = 36\text{ V}

Tahmini Süre:1m 0s

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