Answer:
- v1 = 8V; v2=12V
- i1=9/7A, i2=13/14A, i3=5/14A, v1=18/7V, v2=52/7V, v3=10/7V
Step-by-step explanation:
You want the voltages in each circuit, and also the currents in the second circuit.
1. Voltage divider
In this series circuit, the voltage is divided in proportion to the resistance.
v1 = 2/5(20V) = 8V
v2 = 3/5(20V) = 12V
2. Current equations
The sum of voltages around a loop is 0, so we can write the equations ...
2·i1 +8·i2 = 10
8·i2 -4·i3 = 6
i1 -i2 -i3 = 0
The attachment shows the calculation of the currents. Those are used to find the corresponding voltages.
(i1, i2, i3) = (9/7, 13/14, 5/14)A
(v1, v2, v3) = (18/7, 52/7, 10/7)V
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Additional comment
A T-circuit as in figure 2 can usually be solved handily by making use of Norton's equivalents for the sources. The left source can be replaced by a 5A current source in parallel with 2Ω. The right source can be replaced by a 1.5A current source in parallel with 4Ω. Then the circuit degenerates to a 6.5A source in parallel with 8/(4+1+2) = 8/7Ω. So, the voltage v2 is ...
v2 = (6.5A)(8/7Ω) = 52/7V
Then {v1, -v3} = {10, 6} -v2 ⇒ (v1, v3) = (18/7, 10/7)
The currents are found by dividing the voltage by the resistance:
{i1, i2, i3} = {18/7, 52/7, 10/7}÷{2, 8, 4} = (9/7, 13/14, 5/14) . . . . as above
Note that these calculations can all be done without the aid of calculator.
Parallel resistors that are multiples of one another can be thought of as some number of resistors in parallel. Here, the 2Ω resistor can be thought of as 4 8Ω resistors in parallel. Similarly, the 4Ω resistor is effectively 2 8Ω resistors in parallel. Thus the parallel combination of 2Ω, 8Ω, and 4Ω is effectively 4+1+2 = 7 8Ω resistors in parallel, or 8/7Ω. No calculator required.