Question

Three resistance of 8,12, and 24 ohms are in parallel. This combination is in series with another parallel of 10,10, and 20 ohms. The whole combination is then connected to a battery of 16.6 volts. If the current through the 20 ohm resistor is 0.4 amperes. Find: a.) The internal resistance of the batter b.) The current through the 12 ohm resistor

Answer 1

Replace each parallel combination by a single equivalent resistor :
R1 = 8 Ohms, R2 = 12 Ohms, R3 = 24 Ohms.
1/Req1 = 1/8 + 1/12 + 1/24.
1/Req1 = 3/24 + 2/24 + 1/24 = 6/24.
Req1 = 24/6 = 4 Ohms.

R4 = 10 Ohms, R5 = 10 Ohms, R6 = 20 Ohms.
1/Req2 = 1/10 + 1/10 + 1/20.
1/Req2 = 2/20 + 2/20 + 1/20 = 5/20,
Req2 = 20/5 = 4 Ohms.

a. V4 = V5 = V6 = I6*R6 = 0.4*20 = 8 Volts.
I4 = I5 = 8/10 = 0.8A.

It = I4+I5+I6 = 0.8 + 0.8 + 0.4 = 2A. = Total current.
V1 = V2 = V3 = It*Req1 = 2 * 4 = 8 Volts.
Vt = 8 + 8 = 16 Volts.
Ri = (E-Vt)/It = (16.6-16)/2 = 0.3 Ohms = Internal resistance.

b. I2 = V2/R2 = 8/12 = 0.667A.

Answer 2

Calculate the total resistance and current in a parallel circuit with 3 resistors of 4, 8, and 16 ohms