You have a 6-volt battery (assumed ideal) and a 1.5-volt flashlight bulb, which is known to draw \(0.5 A\) when the bulb voltage is \(1.5 V\) (see figure below). Design a network of resistors to go between the battery and the bulb to give \(v_s = 1.5 V\) when the bulb is connected, yet ensures that \(v_s\) does not rise above \(2 V\) when the bulb is disconnected.



Hint: use a two-resistor voltage divider to create the voltage for node A. You'll have two unknowns (R1 and R2) which can be determined by solving the two equations for \(v_s\) derived from the constraints above: one involving R1, R2 and Rbulb where \(v_s = 1.5\), and one involving R1 and R2 where \(v_s = 2\).

There are two schematic diagrams below. Please enter the network of resistors you've designed into both diagrams. The top diagram is the model when the bulb is connected; the bottom diagram is the model when the bulb is disconnected.

Run a DC analysis on both diagrams to show that the node labeled "A" has a voltage of approximately \(1.5 V\) in the top diagram and less than \(2 V\) in the bottom adiagram. Submit your results after the DC analyses have been run (so the results of the analyses will be submitted too). Schematic model when bulb is connected: Schematic model when bulb is disconnected: # for a schematic response, submission[i] is the json representation # of the diagram and analysis results for the i-th schematic tag correct = ['incorrect', 'incorrect'] # optimistic default :) def get_dc(json): for element in json: if element[0] == 'dc': return element[1] return None dc_with_bulb = get_dc(submission[0]) if dc_with_bulb: v = dc_with_bulb['A'] if v >= 1.4 and v <= 1.6: # want 1.5 correct[0] = 'correct' dc_without_bulb = get_dc(submission[1]) if dc_without_bulb: v = dc_without_bulb['A'] if v <= 2.1: # want 2 correct[1] = 'correct'