Suppose that the battery has a total current in a series connection of 12.0 Ω and the lamps resistances R₁, R₂ and R₃ are 6.0 Ω, 4.0 Ω and 2.0 Ω, respectively. Compute for (a) the total resistance in a series connection and (b) the voltage drops across each resistor.

Step-by-step explanation:

Let the total current = 12A

If there are there lamps 6.0 Ω, 4.0 Ω and 2.0 Ω in series,

a) The total resistance in series connected circuit will be:

Rt = R₁+R₂+R₃

Rt = 6+4+2

Rt = 12Ω

b) According to ohm's law, the voltage drop is expressed as V = IR

I is the current in each lamps

Note that the same current flows in the resistors for series connection

R are the resistances

For the 6Ω, resistor

V = IR

V = 12(6)

V = 72V

The voltage drop across it is 72V

For the 4Ω, resistor

V = IR

V = 12(4)

V = 48V

The voltage drop across it is 48V

For the 2Ω, resistor

V = IR

V = 12(2)

V = 24V

The voltage drop across it is 24V

-2 is the right

sorry man i accidentally pressed yes my bad

step-by-step explanation: