# Series-parallel circuit confusion?

I have a 24 volt circuit that the first leg of the circuit is in series with a 10ohm resistor then the circuit goes into parallel with a 6ohm on left and 12ohm on right then the circuit goes back into a series with one 8 ohm resistor. Here is were I am lost I know on a series the volts will drop and in parallel the amps will drop. In circuit 24 V, 22 Ohm, 1.0909 Amp. But when the parallel circuit is calculated the amps are not equaling to the total circuit amps? I am getting a volt drop through the first 10ohm of 10.0909 Vd so as the remaining 13.0909 volts run through the parallel the amps are not dropping. If someone can give me info on were my calculations might be wrong or a different way to calculate.

Update:

Ok here is a drawing of the circut.

http://www.filedropper.com/circuit

Relevance

You did not show your calculations so I am not sure what you are asking or where you may have made a mistake. I will try to show what I think you might be asking.You have calculated the correct total resistance and the correct total current. Current through a branch of a parallel circuit is inversely proportional to the resistance in that branch.

Thus current through the 6 Ohm resistor equals 1.0909A x ( 12 Ohm)/ (6 Ohm + 12 Ohm) = .7273 Amps.

The current through the 12 Ohm resistor equals 1.0909A x (6 Ohm) / (6 Ohm + 12 Ohm) = .3636 Amps.

Of course the current through the 6 Ohm resistor plus the current. through the12 Ohm resistor must equal the total current. So .7272A + .3636A = 1.0908A ( I lost .0001A somewhere in rounding off)

You were not correct in calculating the Voltage drop across the first 10 Ohm resistor as 10.0909 Volts. This Voltage should be 1.0909A x 10 Ohms = 10.909 Volts. Now you have 24V - 10.909V = 13.091V remaining to be dropped across the remainder of the circuit. Six Ohms in parallel with 12 Ohms = 4 Ohms. So now the resistance of the remainder of the circuit equals 4 Ohms + 8 Ohms = 12 Ohms. The remaining Voltage divided by the remaining resistance should equal the total current. 13.091V / 12 Ohms = 1.0909 Amps

The Voltage drops around the circuit are as follows;

V of 10 Ohm resistor = [(10 Ohm) / (10 + 4 +8) Ohms] x 24V = 10.909V.

V of 6 Ohms in parallel with 12 Ohms = V of 4 Ohms = [(4 Ohms) / (10 + 4 + 8) Ohms] x 24V = 4.364V.

V of 8 Ohm resistor = [(8 Ohms) / ( 10 + 4 + 8)] Ohms x 24V = 8.727V

Total Voltage equals the sum of the drops = 10.909V + 4.364V + 8.727V = 24 Volts.

• Anonymous
5 years ago

I can completely understand your frustration on this circuit. Even though this is technically a parallel circuit, it is also close to a bridge circuit. If you add up the resistances that are in each branch, then yes the voltage would be equal. However, it's asking for the difference between P and Q which are before those secondary resistors are "considered". If you simplify the circuit, it will make a lot more sense. It's not really 10 ohm and 2 ohm in parallel it's 15 and 20. You are right the current doesn't change, but the voltage going through the resistors do. Think about the circuit like a ladder circuit instead. I swear it will make sense after that.

• Not sure about the circuit Is the last 8 ohm resistor in parallel with the other two or is the parallel combination of the 6 & 12 ohm in series with the 10 ohm & 8 ohm.

• 6 and 12 in parallel are 6*12/18 = 4 ohms. total resistance is 4+10+8 = 22 ohms.

Current is 24/22 = 1.09 amps

V10 = 10.9 volts (voltage across the 10 ohm resistor)

V8 = 8.73 volts (voltage across the 8 ohm resistor)

V4 = 4.36 volts (voltage across the parallel two)

and they add up to 24 volts.

Your comments I have trouble understanding, so I'll leave it at that.

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