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Zorluk: OrtaElectricity and Magnetism

An emergency backup lighting system connects three identical 15 Ω15\,\Omega indicator lamps in series across a 45 V45\text{ V} direct-current power source. What is the electric current flowing through each lamp?

  1. A
    0.33 A0.33\text{ A}
  2. B
    3.0 A3.0\text{ A}
  3. 1.0 A1.0\text{ A}Cevap
  4. D
    9.0 A9.0\text{ A}

Cevap

1.0 A1.0\text{ A}
In a series circuit, the total electrical resistance is the sum of all individual resistances in the path: Rtotal=15 Ω+15 Ω+15 Ω=45 ΩR_{\text{total}} = 15\,\Omega + 15\,\Omega + 15\,\Omega = 45\,\Omega. Applying Ohm's law (I=VRI = \frac{V}{R}), the current drawn from the 45 V45\text{ V} source is I=45 V45 Ω=1.0 AI = \frac{45\text{ V}}{45\,\Omega} = 1.0\text{ A}. In any series circuit, the electric current is identical at every point, meaning exactly 1.0 A1.0\text{ A} flows through each lamp.

Adım Adım Çözüm

1
Calculate the total equivalent resistance of the series circuit.
Rtotal=15 Ω+15 Ω+15 Ω=45 ΩR_{\text{total}} = 15\,\Omega + 15\,\Omega + 15\,\Omega = 45\,\Omega
In a series circuit, the total electrical resistance equals the direct sum of all individual resistances.
2
Apply Ohm's law to determine the total circuit current.
I=VRtotal=45 V45 Ω=1.0 AI = \frac{V}{R_{\text{total}}} = \frac{45\text{ V}}{45\,\Omega} = 1.0\text{ A}
Current is calculated by dividing total source voltage by the total resistance of the circuit.
3
Determine the current flowing through each individual lamp.
Ilamp=1.0 AI_{\text{lamp}} = 1.0\text{ A}
Because there is only one continuous path for electrical charge in a series circuit, the current is uniform through all components.

Anahtar Kavram

Ohm's law and total resistance in series circuits
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