An electronics technician is testing a DC bias circuit containing a fixed resistor connect… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician is testing a DC bias circuit containing a fixed resistor connected to an adjustable DC power supply. When the supply voltage is set to $28\text{ V}$ , the circuit ammeter measures a current of $3.5\text{ mA}$ . To increase the current flow through the resistor to $11\text{ mA}$ , what must the power supply voltage be set to?

A
$60\text{ V}$
B
$8.8\text{ V}$
$88\text{ V}$ Cevap
D
$22\text{ V}$

Cevap

The power supply voltage must be set to

88\text{ V}

The power supply must be set to

88\text{ V}

. First, determine the component's constant resistance:

R = \frac{28\text{ V}}{3.5\text{ mA}} = 8\text{ k}\Omega

. Next, apply Ohm's Law to find the voltage required for the target current:

V = 11\text{ mA} \times 8\text{ k}\Omega = 88\text{ V}

. Alternatively, using direct proportionality:

V_2 = 28\text{ V} \times \left(\frac{11\text{ mA}}{3.5\text{ mA}}\right) = 88\text{ V}

Adım Adım Çözüm

Calculate the fixed resistance using the initial voltage and current measurements.

R = \frac{V_1}{I_1} = \frac{28\text{ V}}{3.5\text{ mA}} = \frac{28\text{ V}}{0.0035\text{ A}} = 8{}000\,\Omega = 8\text{ k}\Omega

Ohm's Law defines resistance as the ratio of voltage to current (

R = \frac{V}{I}

Calculate the new voltage required to produce

11\text{ mA}

through the

8\text{ k}\Omega

resistor.

V_2 = I_2 \times R = 0.011\text{ A} \times 8{}000\,\Omega = 88\text{ V}

Applying Ohm's Law in the form

V = I \times R

determines the potential difference needed for the desired current.

Anahtar Kavram

Ohm's Law defines the fundamental linear relationship between voltage, current, and resistance in a DC circuit (

V = I \times R

). In a circuit with constant resistance, voltage is directly proportional to current.

Alternatif Yöntem

Because resistance is constant, voltage is directly proportional to current:

V_2 = V_1 \times \frac{I_2}{I_1} = 28\text{ V} \times \frac{11\text{ mA}}{3.5\text{ mA}} = 28 \times \frac{22}{7} = 4 \times 22 = 88\text{ V}

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