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Zorluk: ZorVoltage and Electromotive Force (EMF)

In an electronic flash unit, a high-voltage trigger circuit transfers 0.72 J0.72\text{ J} of electrical energy by moving a charge of 1.6 mC1.6\text{ mC} across a gas-discharge tube. What is the electromotive force (voltage) developed across the tube during this discharge?

  1. A
    0.45 V0.45\text{ V}
  2. B
    1.15 V1.15\text{ V}
  3. 450 V450\text{ V}Cevap
  4. D
    2.22 V2.22\text{ V}

Cevap

450 V450\text{ V}
Electromotive force (EMF), or voltage, is defined as the amount of work or energy transferred per unit of electric charge, expressed by the formula V=WQV = \frac{W}{Q}, where 1 volt=1 joule per coulomb1\text{ volt} = 1\text{ joule per coulomb}. Converting the charge from millicoulombs to coulombs gives 1.6 mC=0.0016 C1.6\text{ mC} = 0.0016\text{ C}. Dividing the total transferred energy (0.72 J0.72\text{ J}) by the charge (0.0016 C0.0016\text{ C}) produces 0.720.0016=450 V\frac{0.72}{0.0016} = 450\text{ V}.

Adım Adım Çözüm

1
Identify the given physical quantities and their base units.
Energy W=0.72 JW = 0.72\text{ J} and charge Q=1.6 mCQ = 1.6\text{ mC}.
Standard electrical calculations require values expressed in base SI units.
2
Convert the electric charge from millicoulombs to coulombs.
Q=1.6×10−3 C=0.0016 CQ = 1.6 \times 10^{-3}\text{ C} = 0.0016\text{ C}.
The prefix milli- represents 10−310^{-3}, so 1.6 mC=0.0016 C1.6\text{ mC} = 0.0016\text{ C}.
3
Apply the definition formula of electromotive force and voltage (V=WQV = \frac{W}{Q}).
V=0.72 J0.0016 C=450 VV = \frac{0.72\text{ J}}{0.0016\text{ C}} = 450\text{ V}.
One volt is defined as one joule of energy transferred per coulomb of charge.

Anahtar Kavram

Electromotive force (EMF) and voltage represent the potential energy transferred per unit electric charge (1 V=1 J/C1\text{ V} = 1\text{ J/C}).
Tahmini Süre:1m 30s
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