Verified answer

a)

Because they are in series you can add them together and get Req of 2 ohms.You now have 3 resisters in parallel. 2 are 3.8 ohms and one is 2 ohms.

1/Req = 1/2+1/3.8+ 1/3.8

Req = 1.0 ohms

You can now add that Req tot he 3 ohm and 2 ohm in series and get a 6.0 ohm.

You add that in parallel to the 6 ohm;

1/Req = 1/ 6+ 1/6.0

Req = 3 ohms

3) Current out of the battery = 11 v/ 3 ohms = 3.67 ohms

Recall the Req in parallel with the 6 ohm resister is 6.o ohms also. There fore half or 1.83 amps will go through the 2 ohm resister. That is the same amount of current that goes through the 3 ohm at the bottom.

2) The amount of current that goes through one of the 3.8 ohm resisters is proportional to the resistance of all the other resisters in that part of the circuit.

I = 1.83 amps *((3.8+2.0/(3.8+2+3.8)) = 1.11 amps

Use V=ir to fine the voltage and you get 4.2 volts

you're on a roll! For parallel resistances, the reciprocal of the completed resistance is the sum of the reciprocals of the guy resistances. on your final occasion, you in truth calculated the reciprocals with the help of finding the present by way of each and each resistor. Now, basically upload the currents up and divide that sum into the 12 V furnish to discover the parallel resistance.

you're on a roll! For parallel resistances, the reciprocal of the entire resistance is the sum of the reciprocals of the guy resistances. on your final occasion, you in reality calculated the reciprocals via looking the present by using each and each resistor. Now, only upload the currents up and divide that sum into the 12 V grant to discover the parallel resistance.