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Difficulty: HardOhm's Law

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 V28\text{ V}, the circuit ammeter measures a current of 3.5 mA3.5\text{ mA}. To increase the current flow through the resistor to 11 mA11\text{ mA}, what must the power supply voltage be set to?

  1. A
    60 V60\text{ V}
  2. B
    8.8 V8.8\text{ V}
  3. 88 V88\text{ V}Answer
  4. D
    22 V22\text{ V}

Answer

The power supply voltage must be set to 88 V88\text{ V}.
The power supply must be set to 88 V88\text{ V}. First, determine the component's constant resistance: R=28 V3.5 mA=8 kΩR = \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 mA×8 kΩ=88 VV = 11\text{ mA} \times 8\text{ k}\Omega = 88\text{ V}. Alternatively, using direct proportionality: V2=28 V×(11 mA3.5 mA)=88 VV_2 = 28\text{ V} \times \left(\frac{11\text{ mA}}{3.5\text{ mA}}\right) = 88\text{ V}.

Step-by-Step Solution

1
Calculate the fixed resistance using the initial voltage and current measurements.
R=V1I1=28 V3.5 mA=28 V0.0035 A=8000 Ω=8 kΩR = \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=VIR = \frac{V}{I}).
2
Calculate the new voltage required to produce 11 mA11\text{ mA} through the 8 kΩ8\text{ k}\Omega resistor.
V2=I2×R=0.011 A×8000 Ω=88 VV_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×RV = I \times R determines the potential difference needed for the desired current.

Key Concept

Ohm's Law defines the fundamental linear relationship between voltage, current, and resistance in a DC circuit (V=I×RV = I \times R). In a circuit with constant resistance, voltage is directly proportional to current.

Alternative Method

Because resistance is constant, voltage is directly proportional to current: V2=V1×I2I1=28 V×11 mA3.5 mA=28×227=4×22=88 VV_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}.
Estimated Time:45s
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Topics

Ohm's Law

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