Asked by Rony

Resistors R1, R2, and R3 have resistances of 15.0 Ohms, 9.0 Ohms and 8.0 Ohms respectively. R1 and R2 are connected in series, and their combination is in parallel with R3 to form a load across a 6.0-V battery.
-What is the total resistance of the load?
-What is the current in R3
-What is the potential difference across R2?
Answered by drwls
First get the effective resistance of R1 and R2 in series. It is 24 ohms. That in parallel with 8 ohms has an overall circuit resistance Reff geiven by
1/Reff = 1/24 + 1/8 = 4/24 = 1/6
Therefore Reff = 6 ohms.

The total current going through the circuit (battery) is
I = V/Reff = 1.0 amperes.

The total current going through the R1 and R2 is V/(R1+R2) = 6/24 = 0.25 A
The potential drop across R2 is
0.25A*9 ohms = 2.25 V
Answered by Henry
a. Req = (R1+R2)R3/(R1+R2+R3) = (15+9)8/(15+9+8) = 6 Ohms.

I = E/Req = 6/6 = 1A.

b .I2+I3 = 1A.
I2 + 3I2 = 1,
I2 = 1/4 A.
I3 = 1-1/4 = 4/4-1/4 = 3/4 = 0.75A.

c. V2 = I2*R2 = 0.25 * 9 = 2.25 Volts.


Answered by Anonymous
Given:
R1 = 15 ohms
R2 = 20 ohms
R3 = 25 ohms
It = 12A



Solve for:
Rt = ?
Ed1 = R1
Ed2 = R2
Ed3 = R3
Et = ?
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