Download Solutions of problems

Discussion Session on DC circuits
Problem 1 on resistance dependence on T
The resistivity of tungsten increases approximately linearly from 56 n m at 293 K and 1.1   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 , radius r,
and length :
Solve for r to obtain:
P
V2
R
R
V2
P
2

100 V 
R
40 W
R
 250 

 r2
r

R
and the diameter d of the filament is
d 2
Because the resistivity varies
linearly with temperature, we can
use a proportion to find its value at
2500 K:
Solve for 2500 K to obtain:

R
(1)
 2500 K   293 K 2500 K  293 K

3500 K   293 K 3500 K  293 K

 2500K 
2207
3207
2207
3500K  293K   293K
3207
Substitute numerical values and evaluate 2500 K:
 2500 K 
2207
1.1   m  56 n  m   56 n  m  774.5 n  m
3207
Substitute numerical values in
equation (1) and evaluate d:
d 2
774.5 n  m0.5 cm
 250 
 4.44 m
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.
QuickTi me™ and a
TIFF (U ncompressed) decompressor
are needed to see this pi cture.
Picture the Problem Note that when both switches are closed the 50- resistor
is shorted. With both switches open, we can apply Kirchhoff’s loop rule to find the
current I in the 100- resistor. With the switches closed, the 100- 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- 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- 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-
resistor and R:
  300 I  100 I  50 I  0
I

450 

1.5 V
 3.33 mA
450 
100 I100  RI R
I tot  I100  I R
or
I R  I tot  I100
where Itot is the current drawn from the
source when both switches are closed.
Substitute to obtain:
100 I100  RI tot  I100 
or
I100 
Express the current Itot drawn from
the source with both switches
closed:
I tot 
Express the equivalent resistance
when both switches are closed:
Req 
Substitute to obtain:
I tot 
RI tot
R  100 

Req
100  R
R  100 
I100
 300 
1.5 V
100  R
R  100 
Substitute in equation (1) to obtain:
(1)
 300 




R
1 .5 V



R  100   100  R  300  
 R  100 



1.5 V R

400  R  30,000  2
 3.33 mA
Solve for and evaluate R:
R  600 
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?
QuickTime™ and a
TIFF (U ncompressed) decompressor
are needed to see this pi cture.
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:
50 V  I 0 10    I 0 Req  0
50 V
10   Req
Solve for I0:
I0 
Find the equivalent resistance of 15
, 12 , and 15  in parallel:
1
1
1
1



Req 15  12  15 
and
Req  4.62 
Substitute for Req and evaluate I0:
(b) Apply Kirchhoff’s loop rule to
the circuit a long time after the
switch is closed:
I0 
50 V
 3.42 A
10   4.62 
50 V  I  10    I  Req  0
50 V
10   Req
Solve for I:
I 
Find the equivalent resistance of 15
, 12 , and 15  in series:
Req  15   12   15   42 
Substitute for Req and evaluate I:
I 
(c) Using the definition of
capacitance, express the charge on
the capacitors in terms of their final
potential differences:
Q10 F  C10 FV10 F
Apply Kirchhoff’s loop rule to the
loop containing the 15- and 12-
resistors and the 10 F capacitor to
obtain:
Solve for V10 F:
50 V
 0.962 A
10   42 
(1)
and
Q5 F  C5 FV5 F
V10 F  15  I   12  I   0
V10 F  27  I 
(2)
Substitute in equation (1) and
evaluate Q10 F:
Q10 F  C10 F 27  I 
 10 F27  0.962 A 
 260 C
Apply Kirchhoff’s loop rule to the
loop containing the 15- and 12-
resistors and the 5 F capacitor to
obtain:
Solve for V5 F:
Substitute in equation (2) and
evaluate Q5 F:
V5 F  15  I   12  I   0
V5 F  27  I 
Q5 F  C5 F 27  I 
 5 F27  0.962 A 
 130 C
Problem n. 3
A capacitor with capacitance C1 = 100 F, charged with a potential difference V = 1200 V, is
connected through a resistor R = 105  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.
q1
q
 Ri  2
C1
C2
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
1)
