DC Distributor Voltage Drop Calculation Explained

This problem asks us to calculate the maximum voltage drop in a 2-wire DC distributor cable that is uniformly loaded and fed from both ends with equal voltages.

Here's a breakdown of the given information:

• Distributor length: 800 m
• Load distribution: 1 A/m (uniform)
• Resistance of each conductor: 0.05 Ω/km
• Feeding voltage at both ends: 220 V (equal)

Let's perform the calculations step-by-step to find the maximum voltage drop in this DC distributor.

Step 1: Calculate total length in kilometers

The length is given in meters, so we convert it to kilometers:

Length $L = 800 \text{ m} = \frac{800}{1000} \text{ km} = 0.8 \text{ km}$

Step 2: Calculate the total resistance of each conductor

The resistance per kilometer of each conductor is 0.05 Ω/km. The total resistance of one conductor over 0.8 km is:

$R_{conductor} = 0.05 \frac{\Omega}{\text{km}} \times 0.8 \text{ km} = 0.04 \Omega$

Step 3: Calculate the total resistance of the 2-wire distributor cable

A 2-wire distributor has two conductors (go and return). The total resistance of the cable path is twice the resistance of one conductor:

$R_{cable} = 2 \times R_{conductor} = 2 \times 0.04 \Omega = 0.08 \Omega$

Step 4: Calculate the total current drawn by the distributor

The load is uniformly distributed at 1 A/m over the 800 m length:

$I_{total} = \text{Load per meter} \times \text{Length} = 1 \frac{\text{A}}{\text{m}} \times 800 \text{ m} = 800 \text{ A}$

Step 5: Determine the current supplied by each end

When a uniformly loaded distributor is fed from both ends with equal voltages, the total current is shared equally between the two ends due to symmetry. Let $I_A$ and $I_B$ be the currents supplied by the two ends:

$I_A = I_B = \frac{I_{total}}{2} = \frac{800 \text{ A}}{2} = 400 \text{ A}$

Step 6: Locate the point of maximum voltage drop

For a uniformly loaded DC distributor fed from both ends with equal voltages, the point of minimum voltage (and thus maximum voltage drop relative to the feeding points) occurs at the midpoint of the distributor. This is the point where the current flowing from one end meets the current flowing from the other end, resulting in zero net current flow at that specific point.

The midpoint is at a distance of $800 \text{ m} / 2 = 400 \text{ m}$ from either end.

Step 7: Calculate the voltage drop from one end to the midpoint

Let's calculate the voltage drop from one end (say, end A) to the midpoint (400 m away). The current supplied by end A is 400 A. As we move away from end A, the current in the cable decreases linearly due to the uniform load. The current at a distance $x$ meters from end A is $I(x) = I_A - (\text{load per meter} \times x)$.

$I(x) = 400 \text{ A} - (1 \frac{\text{A}}{\text{m}} \times x) = (400 - x) \text{ A}$

The resistance per meter of the 2-wire cable is:

$R_{per\_meter} = \frac{R_{cable}}{\text{Length in meters}} = \frac{0.08 \Omega}{800 \text{ m}} = 0.0001 \frac{\Omega}{\text{m}}$

The voltage drop $dV$ over a small length $dx$ at distance $x$ is $dV = I(x) \times R_{per\_meter} \times dx$.

$dV = (400 - x) \times 0.0001 \times dx$

To find the total voltage drop from end A ($x=0$) to the midpoint ($x=400 \text{ m}$), we integrate this expression:

$V_{drop\_A\_mid} = \int_{0}^{400} (400 - x) \times 0.0001 dx$

$V_{drop\_A\_mid} = 0.0001 \int_{0}^{400} (400 - x) dx$

$V_{drop\_A\_mid} = 0.0001 \left[ 400x - \frac{x^2}{2} \right]_{0}^{400}$

$V_{drop\_A\_mid} = 0.0001 \left[ \left( 400 \times 400 - \frac{400^2}{2} \right) - \left( 400 \times 0 - \frac{0^2}{2} \right) \right]$

$V_{drop\_A\_mid} = 0.0001 \left[ \left( 160000 - \frac{160000}{2} \right) - 0 \right]$

$V_{drop\_A\_mid} = 0.0001 \left[ 160000 - 80000 \right]$

$V_{drop\_A\_mid} = 0.0001 \times 80000$

$V_{drop\_A\_mid} = 8 \text{ V}$

The maximum voltage drop occurs at the midpoint and is 8 V. This represents the drop in voltage from the feeding voltage (220 V) down to the minimum voltage at the midpoint.

The voltage at the midpoint would be $220 \text{ V} - 8 \text{ V} = 212 \text{ V}$. The maximum voltage drop is the difference between the feeding voltage and the minimum voltage point.

The maximum voltage drop is 8 V.