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Since V1 is 10V and the voltage at the tap is 5V, and the current through R1 is 10mA, then R1 = E/I = (10-5)/10m = 500Ω
The current through R2 is I = E/R = 5/5K = 1mA.
The current through the diode& RL must be 10mA-1mA = 9mA (Σ currents into node=0)
The voltage across RL is (5-0.7)=4.3V, therefore RL=E/I = 4.3/9m = 478Ω
btw: your assumption that the forward drop across the diode is 0.700V is a little off. At room temperature, a 1N4148 would have a forward drop of about 0.688V. However, the calculations above are based on 0.7V.
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