Three resistors are connected in series to a 120V generator. The first has a resistance of 50ohms?

Question

the second passes a current of .05A and the third has a voltage drop of 50V across it. Calculate the resistance of the second and the third resistor.

Answer 1

ok, here's how you do it:

First, you know that the current through all the resistors will be 0.05 A since they are in series, so for the third resistor, the resistance will be according to ohms law

R = 50V/0.05A = 1000 Ohm

now that you know two of the resistances of the circuit, you have to get the total resistance of it. You know that the total voltage will be 120 V and the current will be 0.05A then

Rtot = 120V/0.05A = 2400 Ohm

now, you know that Rtot will be the summ of all the resistances

Rtot = R1+R2+R3

then

R2 = Rtot - R1-R3

R2=2400 Ohms- 50 Ohms - 1000 Ohms = 1350 Ohms

Hope this helps.

Answer 2

120 V,

Current is same in a series circuit, so all three resistors see 0.05 amps

Ohm's Law: V = I x R

Overall resistance  = 120 V ÷ 0.05 A = 2400 Ω

R1 = 50 Ω, therefore voltage drop across R1 = 0.05 amps x 50 Ω = 2.5 V

R3 has a voltage drop of 50 V.  50 V ÷ 0.05 A = 1000 Ω

R2 = 2400 Ω - (50 Ω + 1000 Ω) = 1350 Ω

Voltage drop across R2 = 1350 Ω x 0.05 A = 67.5 V

To check the answer, verify total voltage drop:  2.5 V + 67.5 V + 50 V = 120 V

R2 = 1350 Ω

R3 = 1000 Ω

Answer 3

Current is the same in all 3, therefore the current is 50ma.

Third R has a resistance of R = E/I = 50v/50ma  = 1 k ohm

First R has a voltage drop of E = IR = 50ma x 50 = 2500 mv or 2.5 volts.

Voltage on second R is 120 - 50 - 2.5 = 67.5 volts

Second R = E/I = 67.5v/50ma = 1.35 k ohm

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