Discussion Session on DC Circuits

Discussion Session on DC circuits

Problem 1 on resistance dependence on T

The resistivity of tungsten increases approximately linearly from 56 nW × m at 293 K and 1.1 mW × m at 3500 K. Estimate (a) the resistance and (b) the diameter of a tungsten filament used in a 40 W bulb, assuming that the filament temperature is about 2500 K and that a 100 V dc supply is used to power the light bulb. Assume that the length of the filament is constant and equal to 0.5 cm.

Picture the Problem We can use the relationship between the rate at which energy is transformed into heat and light in the filament and the resistance of and potential difference across the filament to estimate the resistance of the filament. The linear dependence of the resistivity on temperature will allow us to find the resistivity of the filament at 2500 K. We can then use the relationship between the resistance of the filament, its resistivity, and cross-sectional area to find its diameter.
(a) Express the wattage of the lightbulb as a function of its resistance R and the voltage V supplied by the source: /
Solve for R to obtain: /
Substitute numerical values and evaluate R: /
(b) Relate the resistance R of the filament to its resistivity r, radius r, and length : /
Solve for r to obtain: /
and the diameter d of the filament is
(1)
Because the resistivity varies linearly with temperature, we can use a proportion to find its value at 2500 K: /
Solve for r2500 K to obtain: /
Substitute numerical values and evaluate r2500 K:
Substitute numerical values in equation (1) and evaluate d: /

Problem 2 on Direct-Current circuits

In the circuit in the figure, the reading of the ammeter is the same with both switches open and both switches closed. Find the resistance R.

Picture the Problem Note that when both switches are closed the 50-W resistor
is shorted. With both switches open, we can apply Kirchhoff’s loop rule to find the current I in the 100-W resistor. With the switches closed, the 100-W resistor and R are in parallel. Hence, the potential difference across them is the same and we can express the current I100 in terms of the current Itot flowing into the parallel branch whose resistance is R, and the resistance of the 100-W resistor. Itot, in turn, depends on the equivalent resistance of the closed-switch circuit, so we can express I100 = I in terms of R and solve for R.
Apply Kirchhoff’s loop rule to a loop around the outside of the circuit with both switches open: /
Solve for I to obtain: /
Relate the potential difference across the 100-W resistor to the potential difference across R when both switches are closed: /
Apply Kirchhoff’s junction rule at the junction to the left of the 100-W resistor and R: /
or
where Itot is the current drawn from the
source when both switches are closed.
Substitute to obtain: /
or
(1)
Express the current Itot drawn from the source with both switches closed: /
Express the equivalent resistance when both switches are closed: /
Substitute to obtain: /
Substitute in equation (1) to obtain: /
Solve for and evaluate R: /

Remarks: Note that we can also obtain the result in the third step by applying Kirchhoff’s loop rule to the parallel branch of the circuit.

Problems on RC Circuits

Problem 1

The capacitors in the figure are initially uncharged. A) What is the initial value of the battery current when switch S is closed? B) What is the battery current after a long time? C) What are the final charges on the capacitors?

Picture the Problem When the switch is closed, the initial potential differences across the capacitors are zero (they have no charge) and the resistors in the bridge portion of the circuit are in parallel. When a long time has passed, the current through the capacitors will be zero and the resistors will be in series. In both cases, the application of Kirchhoff’s loop rule to the entire circuit will yield the current in the circuit. To find the final charges on the capacitors we can use the definition of capacitance and apply Kirchhoff’s loop rule to the loops containing two resistors and a capacitor to find the potential differences across the capacitors.
(a) Apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed: /
Solve for I0: /
Find the equivalent resistance of 15 W, 12 W, and 15 W in parallel: /
and
Substitute for Req and evaluate I0: /
(b) Apply Kirchhoff’s loop rule to the circuit a long time after the switch is closed: /
Solve for I¥: /
Find the equivalent resistance of 15 W, 12 W, and 15 W in series: /
Substitute for Req and evaluate I¥: /
(c) Using the definition of capacitance, express the charge on the capacitors in terms of their final potential differences: / (1)
and
(2)
Apply Kirchhoff’s loop rule to the loop containing the 15-W and 12-W resistors and the 10 mF capacitor to obtain: /
Solve for V10 mF: /
Substitute in equation (1) and evaluate Q10 mF: /
Apply Kirchhoff’s loop rule to the loop containing the 15-W and 12-W resistors and the 5 mF capacitor to obtain: /
Solve for V5 mF: /
Substitute in equation (2) and evaluate Q5 mF: /

Q

Problem n. 3

A capacitor with capacitance C1 = 100 mF, charged with a potential difference DV = 1200 V, is connected through a resistor R = 105 W to a second capacitor, initially uncharged, with the same (C1 = C2). Calculate:

1)  the final value of the potential difference between the plates of the first capacitor;

2)  the energy dissipated through R during this process;

3)  the time it takes to have that the potential differente on the second capacitor is 30% of its final value.

1)

In the final configuration the charge on the 2 capacitors is the same and so the voltage

V = 600 V

2) 36 J

3) 1.78 s