Question

Find the value of required resistor across AB of the following circuit to transfer maximum power?

• A. \(500\Omega\)
• B. \(800\Omega\)
• C. \(442.8\Omega\)
• D. \(262.5\Omega\)

Hint:

Maximum Power Transfer Theorem: Load resistance \(R_L\) should be equal to the Thevenin resistance \(R_{Th}\) of the source circuit.
To find \(R_{Th}\), deactivate independent sources (voltage sources \(\rightarrow\) short circuit, current sources \(\rightarrow\) open circuit) and calculate the equivalent resistance from the load terminals.

Answer 1

The Correct Option is D

This is Lagrange's linear partial differential equation \(Pp + Qq = R\), where \(p = \partial z / \partial x\) and \(q = \partial z / \partial y\).

Here, \(P = mz-ny\), \(Q = nx-lz\), \(R = ly-mx\).

The auxiliary equations are \(\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}\): 

\[ \frac{dx}{mz-ny} = \frac{dy}{nx-lz} = \frac{dz}{ly-mx} \]

Choose multipliers \(x, y, z\): Each fraction is equal to \(\frac{x dx + y dy + z dz}{x(mz-ny) + y(nx-lz) + z(ly-mx)}\).

Denominator = \(mzx - nxy + nxy - lyz + lyz - mzx = 0\).

Thus, \(x dx + y dy + z dz = 0\).

Integrating gives \(\frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2} = C_1'\), or \(x^2+y^2+z^2 = c_1\). Let \(u = x^2+y^2+z^2\).

Choose multipliers \(l, m, n\): Each fraction is equal to \(\frac{l dx + m dy + n dz}{l(mz-ny) + m(nx-lz) + n(ly-mx)}\).

Denominator = \(lmz - lny + mnx - mlz + nly - nmx = 0\).

Thus, \(l dx + m dy + n dz = 0\).

Integrating gives \(lx + my + nz = c_2\). Let \(v = lx+my+nz\).

The complete solution is \(f(u,v)=0\), or \(u = \phi(v)\).

So, \(x^2+y^2+z^2 = f(lx+my+nz)\).

\[ \boxed{x^2+y^2+z^2 = f(lx+my+nz)} \]