The distance between two bridges on the same plane is 150 m. The cross-section of the line conductor is 1.125 cm² and the weight per 100 m is 120 kg. Determine the sag for a breaking stress of 4220 kg/cm² and a safety factor of 5. Assume that the weight per square meter is 90 kg at atmospheric pressure.

একই সমতলে অবস্থিত দুটি পুলের মধ্যবর্তী দূরত্ব 150 মিটার। লাইন পরিবাহীর প্রস্থচ্ছেদ ১.১২৫ বর্গ সেন্টিমিটার এবং প্রতি 100 মিটারে ওজন ১২০ কেজি। ব্রেকিং স্ট্রেস ৪২২০ kg/cm² এবং সেফটি ফ্যাক্টর ৫ জন্য Sag নির্ণয় কর? মনে কর প্রতি বায়ুমণ্ডলীয় চাপে প্রতি বর্গ মিটারের ওজন ৯০ কেজি।

Step-by-step Solution:

Step 1: Calculate Thevenin's Equivalent Voltage (\(V_{th}\))

Terminals X-Y are open-circuited, so no current flows through the branch containing the \(10\,\Omega\) resistor and the \(X_C = 10\,\Omega\) capacitor connected to terminal X.

Therefore, the open-circuit voltage \(V_{th}\) across terminals X-Y is equal to the voltage drop across the parallel branch \((3 + j14)\,\Omega\).

Using the Voltage Divider Rule:

\(V_{th} = V_s \times \frac{3 + j14}{-j10 + (3 + j14)}\)

\(V_{th} = 100 \times \frac{3 + j14}{3 + j4}\)

Multiplying the numerator and denominator by the complex conjugate of the denominator \((3 - j4)\):

\(V_{th} = 100 \times \frac{(3 + j14)(3 - j4)}{3^2 + 4^2}\)

\(V_{th} = 100 \times \frac{9 - j12 + j42 + 56}{25}\)

\(V_{th} = 100 \times \frac{65 + j30}{25}\)

\(V_{th} = 4 \times (65 + j30) = 260 + j120\text{ V}\)

In polar form:

\(|V_{th}| = \sqrt{260^2 + 120^2} = 286.36\text{ V}\)

\(\theta = \tan^{-1}\left(\frac{120}{260}\right) = 24.78^\circ\)

\(V_{th} = 286.36 \angle 24.78^\circ\text{ V}\)

Step 2: Calculate Thevenin's Equivalent Impedance (\(Z_{th}\))

To find the equivalent impedance \(Z_{th}\), deactivate the independent voltage source by replacing it with a short circuit.

Looking back into terminals X-Y, the \(-j10\,\Omega\) capacitor is in parallel with the \((3 + j14)\,\Omega\) branch, and this combination is in series with the \((10 - j10)\,\Omega\) branch leading to terminal X.

\(Z_{th} = (10 - j10) + \left[ (-j10) \parallel (3 + j14) \right]\)

Calculating the parallel portion (\(Z_p\)):

\(Z_p = \frac{-j10 \times (3 + j14)}{-j10 + 3 + j14} = \frac{140 - j30}{3 + j4}\)

Multiplying by the conjugate \((3 - j4)\):

\(Z_p = \frac{(140 - j30)(3 - j4)}{3^2 + 4^2} = \frac{420 - j560 - j90 - 120}{25} = \frac{300 - j650}{25} = 12 - j26\,\Omega\)

Now, substitute \(Z_p\) back to calculate \(Z_{th}\):

\(Z_{th} = (10 - j10) + (12 - j26) = 22 - j36\,\Omega\)

In polar form:

\(|Z_{th}| = \sqrt{22^2 + (-36)^2} = 42.19\,\Omega\)

\(\theta = \tan^{-1}\left(\frac{-36}{22}\right) = -58.57^\circ\)

\(Z_{th} = 42.19 \angle -58.57^\circ\,\Omega\)

Final Answer:

The Thevenin's equivalent circuit across terminal X-Y consists of:

  
• Thevenin's Voltage (\(V_{th}\)): \(260 + j120\text{ V}\) or \(286.36 \angle 24.78^\circ\text{ V}\)
  
• Thevenin's Impedance (\(Z_{th}\)): \(22 - j36\,\Omega\) or \(42.19 \angle -58.57^\circ\,\Omega\)

Maximum Power Transfer Theorem :

In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a power source with internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

For t<0, the switch is closed.

Capacitor acts as open to DC.

Voltage across the capacitor,

     v(0^-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0^-)= v(0) = 8VAt t>0, The capacitor is discharging.R_th = (12||4) = 3 ohmC= 1/6 FTime constant, Π = R_th×C = 0.5 So, v(t) = v(0)e^-t/Π  = 8e^-t/0.5 =8e^-2t V (Ans.)