Answer:
Explanation:
You want the equivalent resistance and the total current in the given series-parallel circuit.
Series
The equivalent resistance offered by two resistors in series is the sum of their resistance values.
Here, there are four circuit branches that consist of two resistors in series. Working from the left side of the diagram, these are ...
- 1 Ω + 3 Ω = 4 Ω
- 4 Ω + 2 Ω = 6 Ω
- 3 Ω + 2 Ω = 5 Ω
- 1 Ω + 3 Ω = 5 Ω
Parallel
The equivalent resistance offered by two resistors in parallel is the reciprocal of the sum of the reciprocals of their resistance. Another way to say this is that the equivalent conductance is the sum of the conductances of the parallel branches. (Conductance is the inverse of resistance.)
Here there is a pair of parallel branches on the left side of the diagram, and a similar pair of parallel branches on the right side of the diagram.
The sum of the conductances of the parallel branches on the left is ...
1/4 +1/6 = 10/24 . . . . mhos total conductance
Then the equivalent resistance of that pair of parallel branches is ...
1/(10/24) = 24/10 = 2.4 . . . . ohms effective resistance
On the right side, it's a little easier, since the parallel paths have the same resistance. Both are 5 ohms.
1/5 +1/5 = 2/5 . . . . mhos total conductance
1/(2/5) = 5/2 = 2.5 . . . . ohms effective resistance
You will notice that the equivalent of two identical resistors in parallel is half the value of either one. Similarly, the equivalent of N identical resistors in parallel is 1/N times the value of any one of them.
Series again
You will notice that each of the pairs of parallel branches has a resistor in series with it. On the left there is a 0.6 Ω resistor in series with the 2.4 Ω effective resistance of the parallel branches. The total resistance of that path to the left from the battery is ...
0.6 Ω + 2.4 Ω = 3.0 Ω
On the right, there is a 0.5 Ω resistor in series with the 2.5 Ω effective resistance of the parallel branches. The total resistance of that path to the right from the battery is ...
0.5 Ω + 2.5 Ω = 3.0 Ω
Parallel again
The left branch is in parallel with the right branch, so the total conductance of those two branches is ...
1/3.0 +1/3.0 = 2/3.0 . . . . mhos total conductance
1/(2/3) = 3/2 = 1.5 . . . . ohms effective resistance
This tells us the entire resistor network is equivalent to 1.5 ohms resistance in parallel with the battery.
The equivalent resistance is 1.5 ohms.
Current
The current is found by dividing the circuit voltage by the equivalent resistance:
I = V/R
I = (24 V)/(1.5 Ω) = 16 A
The total current from the battery is 16 amperes.
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Additional comment
Though it may seem like a joke, the (archaic) unit of conductance is "mho" (symbol ℧), which is "ohm" spelled backward. These days, the internationally accepted name of the unit of conductance is siemens (symbol S).
It is often relatively easy to reduce a circuit like this to an equivalent by combining the parallel and series paths in the way we have shown above. Another way to "solve" the circuit is to write and solve equations for node voltages and/or branch currents. In general, we find that to be more tedious. The series/parallel computations can often be done mentally.
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