Problem + Hard 11-1    Source:    Problem 23 and 48 - List of Electric Circuit exercises I - School of Engineering - UFRGS - 2011 - Prof. doctor Valner Brusamarello.

In the circuit of the Figure 11-01.1, when E = 90, I = 15 , the resistor R dissipates maximum power. The voltmeter reads 45 volts and the ammeter reads - 15 A. Determine the reading of the voltmeter and the ammeter when E = 20 volts, I = 10 amps and R_x falling to 1/3 of its previously calculated value.

Solution of the Problem + Hard 11-1   -    Transf. of Sources

Using the problem data, the value of R_x can be calculated. The value of the voltage was given on it, which is the value indicated by the voltmeter (45 volts) and the current which circulates by R_x, which is the value indicated by the ammeter (15 A). Soon:

But the problem says that under these conditions, R dissipates the maximum power. For that to happen, you must have R = R_x. Therefore:

With these data, the value of V₁ can be calculated, directly applying Ohm's law, that is:

The figure below shows the circuit with its calculated values. Note that they have been replaced the two resistors in parallel with the current source I by one of value equal to 3 ohms, which is the result of paralleling the 4 and 12 ohms resistors.

Regarding the circuit shown in Figure 11-01.2, note that the current supplied by the 15 A current source will flow only by the 3 ohm resistor. Thus, over this resistor, a potential difference of 45 volts appears, with the polarity indicated in the figure. As V₁ = 90 volts, so it follows that over R₂, there will be a potential difference of 45 volts (90 - 45 = 45). And since E = 90 volts, obviously over R₁ there will also be a potential difference 45 volts, because the current flowing through R₁ is the same one that circulates through R₂. Therefore:

With the first given situation, it was possible to calculate the parameters of the circuit. For the second situation, it is known that R_x = 1 ohm, E = 20 volts and I = 10 A. The current supplied by the 10 A current source, remains in the same situation as before, that is, it will circulate only through the resistor of 3 ohms. So the voltage source E = 20 volts will distribute its voltage equally between R₁ and R₂. So, the potential difference over R₂ will be 10 volts. And how about the resistor of 3 ohms circulates the current of 10 A, then the potential difference will be 30 volts. This allows calculate V₁.

Now, knowing V₁, it is very easy to calculate the readings the voltmeter and the ammeter. Let initially be ammeter reading.

Note the negative value of A, as the current flows clockwise, leaving the positive pole of the ammeter. To calculate the reading voltmeter, just apply Ohm's law to R_x, or: