5. Direct current circuits
In the circuit of the figure, find which of the emf sources provide or absorb energy and compute the power delivered or absorbed by each.
Using the loops method, with loop currents in mA and clockwise direction:
In the 6 V source, the current is 0.874 mA passing from the positive to the negative electrode. Thus, the source provides power 6×0.874 = 5.24 mW. In the 5 V source, the current is 0.8737 - 0.0885 = 0786 mA, from the negative electrode to the positive; therefore, the source provides power equal to 5×0786 = 3.93 mW.
Determine the power dissipated by each resistor in the circuit and ther power provided by the source. Check that the power supplied by the source is equal to the sum of all powers dissipated in the resistors.
Using the loops method with loop currents in the clockwise direction, the loop currents can be stored in a list I using the following command:
The power dissipated in each resistor is R I2. Since values obtained for the currents have denominators of the order of thousands, the powers can be computed in mW as: 1000 R I2
(%o3) [ 82.13, 62.0, 0.045, 105.8, 151.4 ]
The total power dissipated is then
which is equal to the power supplied by the source: 1000×6×(I3 − I1) = 401.4 mW.
In the circuit shown in the diagram, compute: (a) The initial currents in all resistors and capacitors. (b) The final charges in the capacitors, indicating their polarities.
Initial equivalent circuit:
assuming potential equal to zero in some point, one can compute the potential of all the other points in the diagram. Thus, the voltage in the 150 Ω resistor is 6 V and the current through it is 6/150 = 0.04 A (from left to right) which is the same current in the 82 nF capacitor (from right to left). In the 200 Ω resistor the voltage is 1.5 V and the current is 1.5 / 200 = 0.0075 A (top to bottom). By the nodes law, the current in teh 68 nF capacitor is then 0.04 - 0.0075 = 0.0325 A (top to bottom). In the 1.2 kΩ resistor the current is zero, because the voltage in it is null.
The final equivalent circuit is:
The figure shows the values of the potential in the points where it can be calculated from the values of the sources. Note that the current int eh150 Ω resistor is zero, because it has no path through which to circulate.
To compute the potential V in the figure, one must find the current in the 200 Ω and 1.2 kΩ resistorsK, which are in series: R = 1.5/1400 and multiplying by 1200 Ω gives V = 1200(1.5/1400) = 1.286 V. It is observed then that the charge in the 82 nF capacitor is positive on the plate on the right side. The 68 nF capacitor has negative charge on the plate at the top and the values of the charges in those two capacitors are:
(a) Find the intensity and direction of the initial current in the capacitor. (b) Find the final charge in the capacitor and indicate its polarity.
(a) The equivalent circuit in the initial state, with the capacitor in short-circuit is the following
Using the loops method, with 3 loop currents in the counterclockwise direction, the system of equations of the circuit is then
The solution of this system is = 0.00824, = 0.00346 and = 0.00369. The current through the capacitor is , ie 0.00478 A, upwards.
(b) The equivalent circuit in the final stage, with the capacitor as an open switch, is as follows
which is equivalent to the following two simple circuits:
In the circuit on the right side, the current is equal to
and the voltage resistance of 329.4 Ω is:
Thus, the current in the 800 Ω resistor on the circuit on the left side, is equal to (from left to right):
And in the initial circuit, the potential difference between the two points where the capacitor is connected is equal to,
The positive result indicates that the charge is positive in the lower plate and negative in the plate above. Finally, the charge in the capacitor is obtained from the voltage