Question

You are designing a circuit to power a night light, and want to keep the light level as low as possible. The wire you use can handle up to 3 Amps of current; higher than that and the circuit will shut down automatically. If the power supply is 120 V, what is the smallest number of 2.3 Ohm light bulbs you can put in series in the circuit without exceeding the maximum current?

Answer 1

Total 20 number of resistors are required to connect in series

Step-by-step explanation:

As we know by Ohm's law that potential difference across the resistor is the product of resistance and current flowing in it

So we have

`V = i R`

here we know that

V = 120 V

i = 3 A

so we have

`R = (120)/(3) = 40 ohm`

now we know that all the resistors are connected in series and each resistor is of 2 ohm

so total N number of resistors are required

`2 N = 40`

`N = 20`