Delaying a 5 volt power off with an RC circuit
Let's assume for simplicity that the current is constant 600mA and the voltage is dropping from 5V to 4V during 5 seconds. The total charge over 5 sec that is drawn from the capacitor is 600mA * 5 = 3c. The initial charge of a capacitor C is 5V * C. The final charge is 4V * C. The difference is (5V-4V)C = 3. I.e. C=3F. It's a damn large capacitor.