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Zorluk: OrtaSeries Circuits

A diesel generator preheat circuit is powered by a 24 V24\text{ V} DC source connected in series with a ballast resistor R1=1.8 ΩR_1 = 1.8\ \Omega and two identical glow plugs (R2R_2 and R3R_3). If an ammeter in the circuit measures a steady current of 4.0 A4.0\text{ A}, what is the resistance of a single glow plug in ohms (Ω\Omega)?

Cevap: 2.1 Ω

Cevap

The resistance of a single glow plug is 2.1 Ω2.1\ \Omega.
Applying Ohm's Law gives a total series loop resistance of Rtotal=24 V/4.0 A=6.0 ΩR_{\text{total}} = 24\text{ V} / 4.0\text{ A} = 6.0\ \Omega. In a series circuit, total resistance is the sum of all individual resistances (Rtotal=R1+R2+R3R_{\text{total}} = R_1 + R_2 + R_3). Subtracting the ballast resistor (1.8 Ω1.8\ \Omega) leaves 4.2 Ω4.2\ \Omega across the two identical glow plugs. Dividing 4.2 Ω4.2\ \Omega equally between them gives 2.1 Ω2.1\ \Omega per glow plug.

Adım Adım Çözüm

1
Calculate the total circuit resistance using Ohm's Law.
Rtotal=VtotalI=24 V4.0 A=6.0 ΩR_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{24\text{ V}}{4.0\text{ A}} = 6.0\ \Omega
In a closed DC loop, total resistance equals total source voltage divided by total current.
2
Subtract the known ballast resistance from the total resistance to determine the combined resistance of both glow plugs.
R2+R3=Rtotal−R1=6.0 Ω−1.8 Ω=4.2 ΩR_2 + R_3 = R_{\text{total}} - R_1 = 6.0\ \Omega - 1.8\ \Omega = 4.2\ \Omega
Resistances in a series circuit are directly additive (Rtotal=R1+R2+R3R_{\text{total}} = R_1 + R_2 + R_3).
3
Divide the combined glow plug resistance equally between the two identical units.
Rplug=4.2 Ω2=2.1 ΩR_{\text{plug}} = \frac{4.2\ \Omega}{2} = 2.1\ \Omega
Because the two glow plugs are identical, each provides exactly half of the remaining series resistance.

Anahtar Kavram

Series Total Resistance and Ohm's Law
Tahmini Süre:1m 0s
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Series Circuits

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