Right option is (c) 4 Ω
The explanation is: Maximum power is transferred to RL when the load resistance equals the Thevenin resistance of the circuit.
RL = RTH = \(\frac{V_{OC}}{I_{SC}} \)
Due to open-circuit, VOC = 100 V; ISC = I1 + I2
Applying KVL in lower loop, 100 – 8I1 = 0
Or, I1 = \(\frac{100}{8} = \frac{25}{2}\)
And VX = -4I1 = -4 × \(\frac{25}{2}\) = -50V
KVL in upper loop, 100 + VX – 4I2 = 0
I2 = \(\frac{100-50}{4} = \frac{25}{2}\)
Hence, ISC = I1 + I2 = \(\frac{25}{2} + \frac{25}{2}\) = 25
RTH = \(\frac{V_{OC}}{I_{SC}} = \frac{100}{25}\) = 4 Ω
RL = RTH = 4 Ω.