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Difficulty: Very hardSeries Circuits

Analyze the series circuit parameters under normal operation and after a short-circuit fault by filling in the correct numerical values (use decimal format without units, e.g., '0.5' or '10').

Answer:An aviation warning beacon is powered by a 120 V120\text{ V} DC supply connected in series with three resistive components: R1=100 ΩR_1 = 100\ \Omega, R2=150 ΩR_2 = 150\ \Omega, and R3=50 ΩR_3 = 50\ \Omega. Under normal operation, the total current flowing through the circuit is 【0.4】 A and the voltage drop across R1R_1 is 【40】 V. If component R2R_2 experiences a catastrophic failure and completely shorts out (its internal resistance drops to 0 Ω0\ \Omega), the equivalent resistance of the entire circuit drops to 【150】 Ω\Omega, causing the voltage drop across R3R_3 to become 【40】 V.

Answer

Under normal operation, the current is 0.4 A, and the voltage drop across R1 is 40 V. After R2 shorts out, the total resistance drops to 150 ohms, and the increased current causes the voltage drop across R3 to become 40 V.
In a series circuit, total resistance is the linear sum of all resistive components. Current remains uniform throughout the entire loop and is determined by the total voltage divided by the total resistance. When a component fails via a short circuit, it effectively acts as a direct wire (zero ohms). This absence of resistance lowers the total circuit resistance, which consequently forces a higher current to flow from the constant voltage source. The increased current then results in proportionately higher voltage drops across the surviving resistors.

Step-by-Step Solution

1
Calculate the total equivalent resistance under normal operating conditions.
Rtotal=R1+R2+R3=100+150+50=300 ΩR_{total} = R_1 + R_2 + R_3 = 100 + 150 + 50 = 300\ \Omega
In a series circuit, the total resistance is the sum of all individual resistances.
2
Use Ohm's Law to find the total current flowing through the normal circuit.
Itotal=VtotalRtotal=120 V300 Ω=0.4 AI_{total} = \frac{V_{total}}{R_{total}} = \frac{120\text{ V}}{300\ \Omega} = 0.4\text{ A} (This is blank 1).
Total current is equal to the source voltage divided by the total equivalent resistance.
3
Calculate the voltage drop across R1R_1 under normal conditions.
VR1=Itotal×R1=0.4 A×100 Ω=40 VV_{R1} = I_{total} \times R_1 = 0.4\text{ A} \times 100\ \Omega = 40\text{ V} (This is blank 2).
The voltage drop across any single component in a series circuit is the total current multiplied by that component's resistance.
4
Determine the new total resistance after R2R_2 shorts out.
Rnew_total=100+0+50=150 ΩR_{new\_total} = 100 + 0 + 50 = 150\ \Omega (This is blank 3).
A shorted component has 0 Ω0\ \Omega resistance. The circuit remains closed, but the total resistance decreases.
5
Calculate the new total current and the resulting new voltage drop across R3R_3.
Inew=120 V150 Ω=0.8 AI_{new} = \frac{120\text{ V}}{150\ \Omega} = 0.8\text{ A}. Then, VR3=0.8 A×50 Ω=40 VV_{R3} = 0.8\text{ A} \times 50\ \Omega = 40\text{ V} (This is blank 4).
The lowered total resistance causes a higher total current to flow from the same voltage source, which proportionally increases the voltage drop across the remaining functional resistors.

Key Concept

Series Circuit Analysis and Component Fault (Short Circuit) Behavior
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Topics

Series Circuits

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