Problem 10-3 Source:
Problem 2.13 - page 64 - IRWIN, J. David - Book:
Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson Education do Brasil - 2013.
On the circuit show in Figure 10-03.1, determine:
a) The voltage Vo.
b) What power is supplied and/or received by the voltage source.
Solution of the Problem 10-3
Item a
In the Figure 10-03.2 we notice that the circuit that is highlighted in blue presents
two series resistors ( 3 e 9 ohms ) which, in turn, are in parallel
with the resistance of 4 ohms.
Therefore, by performing the calculation, the series results in 12 ohms
and this, in parallel with 4 ohms, results an equivalent resistance of 3 ohms.
In the Figure 10-03.3 we see how the simplified circuit was.
After simplifying the circuit we realize that all resistors are in series.
Therefore, to calculate the Rtotal just add your values. So:
Rtotal = 1 + 2 + 3 = 6 ohms
Now, by applying the Ohm's law, the value of
itotal is:
itotal = 12 / Rtotal = 12 / 6 = 2 A
With the value of itotal, we can calculate Vo,
because by again applying the Ohm's law:
Vo = 1 itotal = 1 2 = 2 volts
Item b
As in the whole circuit we have a single source of voltage it is obvious that this source will be the power supplier for the circuit. Therefore, its power must be negative, indicating that
provides power, or:
Psource = - 12 itotal = - 12 2 = - 24 W
Note that this power is exactly the power consumed by the whole circuit. If we multiply Rtotal by the square of itotal, we get the power dissipated
by the resistors that make up the circuit, or:
Pdissip = Rtotal (itotal)2 = 6 22 = + 24 W
It is known that the algebraic sum of all the powers involved
in the circuit must be equal to ZERO, or: