how do you make a circuit so 1 switch will turn on/off all the lights(3 lights) and a second switch will change the lights from all being the same brightness to all being different brightnesses?

Answers

Answer 1

Answer:

Answer:

The three bulbs can be connected in series to the battery B1, such that when switch SW1 is turned ON all three bulb turns ON with the same brightness. We now have the three resistors R1, R2, and R3 connected in parallel to the lamps LMP1, LMP2, and LMP3 respectively when SW2 turn on. SW2 is a Three Pole Single Throw switch. Please see attachment for the electric circuit diagram.

Explanation:

QUESTION

How do you make a circuit so 1 switch will turn on/off all the lights(3 lights) and a second switch will change the lights from all being the same brightness to all being different brightness?

ANSWER

List of Component Used For The Design Of The Electric Circuit

1. 18Volt Battery ( two 9 Volts batteries connected in series)

2. Three Filament lamps LMP1=LMP2=LMP3= 6V 6W.

3. Three resistors of R1=100Ω, R2=12Ω and R3=3Ω

4. SPST Switch =SW1

5. TPST Switch = SW2

Since the battery is 18V and connected in series to all three lamps, it is capable of delivering 6V per lamp. Each lamp is 6W. So the resistance of the bulb can be determined using Ohm’s law.

V=IR, P=IV = V2/R, R=V2/P

Where I= Current

V= Voltage

R= Resistance

P = Power

Lamp Resistance RL= 62/6 =6Ω

Total resistance of the lamps connected in series Rs= 6+6+6 = 18Ω

The total current through the series resistor combination IS = Battery supplied voltage divide by RS =18/18 =1A

So each lamp will dissipate power PL= I2*R= 1*6 =6W

When switch SW2 is closed, there is now a resistor connected in parallel to each of the lamp which will now reduce the total resistance combination of the lamp and resistors to a values lower than 6Ω

Lamp1 has 100Ω connected in parallel to it to give a total resistance of 5.66Ω

Lamp2 has 12Ω connected in parallel to it to give a total resistance of 4Ω

Lamp3 has 3Ω connected in parallel to it to give a total resistance of 2Ω

The new total resistance of the circuit is now R=5.66+4+2 = 11.66Ω

The new total current flowing through the circuit I=18/11.66 = 1.54A

The power dissipated by each lamp with a new series current of 1.54A can now be recalculated as follows;

Power dissipated by Lamp 1 P1= 1.542*5.66Ω = 13.42W

Power dissipated by Lamp 2 P2= 1.542*4Ω = 9.49W

Power dissipated by Lamp 1 P1= 1.542*2Ω = 4.74W

From the power dissipated by the 3 bulbs, we can see that lamp 3 is bright, lamp 2 is brighter and Lamp 1 is the brightest of all the three lamps.

Answer 2

Answer:
Parallel circuit. On a wire you put the lights and one switch, on the other you put a resistor and another switch. (and the third wire contains the generator)

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