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.
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
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