Control Voltage when the Lamp is Disconnected

This is a simple exercise. The solution is left bottom. The other circuits are helper ones. Probably not a practical circuit - when switched off it still draws 1.33 A. Exercise: You have a 6-volt battery assumed ideal and a 1.5-volt flashlight bulb which is known to draw 0.5A when the bulb voltage is 1.5V see figure below. Design a network of resistors to go between the battery and the bulb to give vs1.5V when the bulb is connected yet ensures that vs does not rise above 2V when the bulb is disconnected. Top: If the lamp draws 0.5A for 1.5V it must consume 0.75W, because P=UI. R = U / I = 3 Ohm Right, the same lamp in series with 9 Ohm resistor - The woltage is split between the resistor and the lamp. Left bottom: When the lamp is disconnected, 3 Ohm and 1 Ohm resistor work like a voltage divider. Vout = Vin * R2/(R1+R2) = 6* 1.5/(4.5) = 6 * 1/3 = 2 V When the circuit is closed Two resistors in parallel 1.5*3 / (1.5+3) = 1 Ohm. With 3 Ohm resistor in series it makes it 4 Ohms equivalent resistence. The current is U/R = 6 V /4 Ohm = 1.5 A. The equivalent circuit is at top right. On the 3 Ohm resistor the voltage drop is 3/4 of 6 V which is 4.5V. This is why 1.5 V comes to the lamp and this is what we want. Kirchhoff's Voltage Law (KVL) : 6 V - 4.5 V - 1.5 V = 0 Kirchhoff’s Current Law (KCL): 1.5 A flows out of the 3 Ohm resistor and is split in ratio 2 : 1 between the 1.5 Ohm resistor and the lamp with 3 Ohms resistence. Learning: solve the open circuit first - it's simpler - just a voltage divider. Start with the most simple equivalent circuit and work out from it. All the euqations must always hold.

published 5 years ago