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Difficulty: Very hardParallel Circuits

An industrial control panel is powered by a 36 V36\text{ V} DC source. The main circuit splits into three parallel branches. Branch one powers a cooling unit that dissipates 108 W108\text{ W} of power. Branch two contains an 18 Ω18\ \Omega limiting resistor. Branch three contains a sensor module with an unknown resistance. If a technician measures the total equivalent resistance of this entire parallel network to be exactly 4 Ω4\ \Omega, what is the resistance of the sensor module?

  1. A
    4 Ω4\ \Omega
  2. 9 Ω9\ \OmegaAnswer
  3. C
    12 Ω12\ \Omega
  4. D
    18 Ω18\ \Omega

Answer

The resistance of the sensor module in the third branch is 9 Ω9\ \Omega.
To find the resistance of a single parallel branch, you need the voltage across it and the current through it. The voltage is the source voltage (36 V36\text{ V}). The total current is 9 A9\text{ A} (36 V/4 Ω36\text{ V} / 4\ \Omega). Subtracting the current from the first branch (108 W/36 V=3 A108\text{ W} / 36\text{ V} = 3\text{ A}) and the second branch (36 V/18 Ω=2 A36\text{ V} / 18\ \Omega = 2\text{ A}) leaves 4 A4\text{ A} flowing through the third branch. Using Ohm's Law (R=V/IR = V/I), 36 V/4 A36\text{ V} / 4\text{ A} equals 9 Ω9\ \Omega.

Step-by-Step Solution

1
Calculate the total current (ItotalI_{total}) drawn by the entire circuit.
Itotal=VRtotal=36 V4 Ω=9 AI_{total} = \frac{V}{R_{total}} = \frac{36\text{ V}}{4\ \Omega} = 9\text{ A}
Ohm's law applies to the circuit as a whole just as it does to individual components.
2
Calculate the current flowing through Branch 1 (I1I_1).
I1=P1V=108 W36 V=3 AI_1 = \frac{P_1}{V} = \frac{108\text{ W}}{36\text{ V}} = 3\text{ A}
In a parallel circuit, the voltage across each branch is equal to the source voltage. We use the power formula (P=V×IP = V \times I) rearranged to solve for current.
3
Calculate the current flowing through Branch 2 (I2I_2).
I2=VR2=36 V18 Ω=2 AI_2 = \frac{V}{R_2} = \frac{36\text{ V}}{18\ \Omega} = 2\text{ A}
Using Ohm's law (I=V/RI = V/R) with the known voltage and resistance for this specific branch.
4
Determine the current flowing through the unknown Branch 3 (I3I_3).
I3=Itotal−(I1+I2)=9 A−(3 A+2 A)=4 AI_3 = I_{total} - (I_1 + I_2) = 9\text{ A} - (3\text{ A} + 2\text{ A}) = 4\text{ A}
Kirchhoff's Current Law states that the total current entering a parallel network must equal the sum of the currents in individual branches.
5
Calculate the resistance of Branch 3 (R3R_3).
R3=VI3=36 V4 A=9 ΩR_3 = \frac{V}{I_3} = \frac{36\text{ V}}{4\text{ A}} = 9\ \Omega
Applying Ohm's law once more to the specific values of the third branch.

Key Concept

Parallel Circuit Current and Power Laws
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Topics

Parallel Circuits

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