Resistors in series and parallel, Kirchhoff's rules
Resistors in series and parallel
Problem:
Four identical light bulbs of resistance R are
connected as shown in the figure.
The battery provides a potential difference V_{0}.
The switches S_{1} and S_{2} can be open and/or
closed in four different combinations: open-open, closed-closed, open-closed,
closed-open.
(a) Consider light bulb A: determine which
switch combinations would produce the brightest and dimmest light in bulb A.
(b) Consider light bulb B: determine which
switch combinations would produce the brightest and dimmest light in bulb B.
Solution:
- Concepts:
Resistors in series and parallel
- Reasoning:
Analyze a simple circuit.
- Details of the calculation:
(a) brightest:
S_{1} closed, S_{2}
open, I_{A} = 2V/(5R).
dimmest:
S_{2} closed, I_{A} = 0.
(b) brightest:
S_{2} closed, S_{1}
open, I_{B} = V/(2R).
dimmest:
S_{1} closed, S_{2}
open, I_{B} = V/(5R).
Problem:
The
circuit shown in the diagram contains an ideal battery and two resistors, R_{1} and R_{2}.
A voltmeter is used to measure the voltage across R_{1},
then across R_{2},
then across the battery.
Its readings are, respectively, 2.0 V; 3.0 V; 6.0 V.
What are the actual voltages across the resistors?
Solution:
- Concepts:
Resistors in series and parallel
- Reasoning:
The voltmeter as has a shunt resistance R. It reads the voltage across
this resistance. When placed in a circuit, it correctly reads the
battery voltage but not the voltages across the resistors of the undisturbed
circuit.
The battery voltage is V = 6 V.
The voltages across R_{1} and R_{2} (without the voltmeter
in the circuit) are
V_{1} = 6V R_{1}/(R_{1} + R_{2}) and V_{2}
= 6V R_{2}/(R_{1} + R_{2}), respectively.
V_{1} = 6V /(1 + R_{2}/R_{1}), V_{2} = 6V /(1 +
R_{1}/R_{2}).
We need to solve for the ratio R_{1}/R_{2}.
- Details of the calculation:
With the voltmeter in the circuit we have:
2V = 6V [R_{1}R/(R_{1} + R)]/[ (R_{1}R/(R_{1} +
R)) + R_{2}] or 2V = 6V/[1 + R_{2}(R_{1} + R)/(R_{1}R)]
3V = 6V [R_{2}R/(R_{2} + R)]/[ (R_{2}R/(R_{2} +
R)) + R_{1}] or 3V = 6V/[1 + R_{1}(R_{2} + R)/(R_{2}R)]
[1 + R_{2}(R_{1} + R)/(R_{1}R)] = 3, R_{2}(R_{1}
+ R)/(R_{1}R) = 2, R_{2}/R + R_{2}/R_{1} = 2,
1/R = 2/R_{2} - 1/R_{1}
[1 + R_{1}(R_{2} + R)/(R_{2}R)] = 2, R_{1}(R_{2}
+ R)/(R_{2}R) = 1, R_{1}/R + R_{1}/R_{2} = 1.
1/R = 1/R_{1} – 1/R_{2
}Therefore 3/R_{2} = 2/R_{1}, R_{1}/R_{2} =
⅔.
V_{1} = 6V *2/5 = 2.4 V, V_{2} = 6V*3/5 = 3.6 V.
Problem:
What is the
resistance of the following network? Each ohmic resistor has resistance R.
Solution:
- Concepts:
Resistors in series and parallel
- Reasoning:
The circuit has enough symmetry so that we can analyze it like a simple
circuit with resistors in series and parallel.
- Details of the calculation:
R_{total} = (3/2)R.
Problem:
(a) Four capacitors are connected as shown in the figure.
C_{1 }= C_{2 }= C_{3 }= C_{4 }= 1
μF.
What is the total capacitance between points A and B?
(b) Five identical 1 Ω resistors are joined and form the four sides of square
and its diagonal. What is the resistance between points A and B?
Solution:
- Concepts:
Capacitors and resistors in series and parallel
- Reasoning:
We are asked to find the effective capacitance and resistance of given
configurations.
- Details of the calculation:
(a) The inverse of the equivalent capacitance for the
sub-circuit with two capacitors C_{3} and C_{4} is
1/C_{eq} = 1/1 + 1/1 = 2, thus C_{eq}= ½. Now, the
equivalent capacitance for the sub-circuit with C_{2}, C_{3},
and C_{4} is C_{eq}'= C + C_{eq} = 1 + ½ = 3/2.
Finally, the equivalent capacitance for the entire circuit is 1/C_{eq}''
= 1/C_{1} + 1/C_{eq}' = 1 + ⅔ = 5/3, thus C_{eq}''=
3/5. The total capacitance is 0.6 uF.
(b) The equivalent resistance of a sub-circuit with R_{3}
and R_{4} is 2. The equivalent resistance of a sub-circuit with R_{3},
R_{4}, and R_{5} is 1/R_{eq} =1/1 + ½ = 3/2, thus R_{eq}
= ⅔. The equivalent resistance of a sub-circuit with R_{2}, R_{3},
R_{4}, and R_{5} is R_{eq}' = R_{2} + Re_{q}
= 1 + ⅔ = 5/3. Finally, the equivalent resistance of the entire circuit is 1/R_{eq}''=
1/1 + 3/5 = 8/5, thus R_{eq}''= 5 /8. The resistance between points A
and B is 5/8 Ω.
Problem:
Find the maximum power of a
heating^{ }element that can be constructed from a piece of wire^{
}that has a resistance of 536 Ω. The element is^{ }to be powered
by a constant voltage of V = 110V. The^{ }current through the wire
cannot exceed 2.0 A.
(a) Assume that you are
allowed to discard a section of the wire,
(b) Assume that you are NOT allowed to discard a section of the wire,
Solution:
- Concepts:
Ohm's law, I = V/R, Power P = IV, resistors in
series and parallel
- Reasoning:
We can cut the wire into N pieces and connect these pieces in parallel to the
power supply.
- Details of the calculation:
Since V is constant, we have to maximize I to maximize P.
For each of the pieces of wire that are connected in parallel we have
I = V/R, I_{max} = 2 A = 110 V /R_{min}, R_{min} = 55
Ω.
(a) We can cut 9 pieces of 55 Ω and discard one piece of 41 Ω.
Then the total current is 9*2 A and the power dissipated
is 1980 W.
(b) We can cut 8 pieces of 55 Ω and 1 piece of 96 Ω. The total current them is
I_{total} = 8*110 V/(55 Ω) + 110 V/(96 Ω). The power is P = VI_{total}
= 1886 W.
But what if we cut some small piece of the 96 Ω wire and added it to one of the
55 Ω wires?
I = V/R, dI = -(V/R^{2})dR
The total change in current through the wires them is dI = -(V/R^{2}_{55Ω})dR
+ (V/R^{2}_{96Ω})dR.
dI is negative, less current flows, less power is dissipated in the heating
element. The maximum power dissipated in the heating
element is 1886W.
Problem:
In the
infinite circuit shown in the diagram, each battery has emf ε and internal
resistance r. Each resistor has resistance 2r. Find the emf and the internal
resistance of the equivalent battery.
Solution:
- Concepts:
An
infinite ladder network
- Reasoning:
Since the ladder is infinite, the current through the equivalent emf and
internal resistance will not
change if an additional section is added to the front of ladder.
- Details of the calculation:
Any combination of batteries and resistances with two terminals can be
replaced by a single voltage source V and a single series resistor R. The
Thevenin voltage V is an ideal voltage source equal to the open circuit
voltage at the terminals. The Thevenin resistance R is the resistance
measured at the terminals with all voltage sources replaced by short
circuits and all current sources replaced by open circuits. Let the
Thevenin voltage of the network be V am let the Thevenin resistance be R.
The network is made of an infinite number of sections.
Let us construct a new two-terminal network by adding another section to the
front of the old two-terminal network.
Let the Thevenin voltage of the new network be V' and let the Thevenin
resistance be R'.
We have
V' = ε +V2r/(2r + R), R' = 2rR/(R + 2r) + r = (3rR + 2r^{2})/(R
+ 2r).
Setting R' = R, V' = V we obtain
R^{2} - rR - 2r^{2} = 0, R = r/2 + ((r/2)^{2} + 2r^{2})^{½}
= 2r.
V = (2r + R)ε/R = 2ε.
The emf of the equivalent battery is 2ε, and the internal resistance is 2r.
Problem:
(a) Calculate the resistance between two points A
and B of the infinite system of resistors.
(b) Calculate the resistance between points A and
B of the cube made of identical resistors r.
Solution:
Problem:
What is the equivalent resistance of the network shown? Each resistor has resistance R.
Solution:
- Concepts:
Resistors in series and parallel
- Reasoning:
No current flows through the center resistor. It can be ignored. Then circuit
has enough symmetry so that we can analyze it like a simple circuit with two
pairs of resistors in series. The pairs are parallel to each other.
- Details of the calculation:
For two resistors in series: R_{eff} = R_{1} + R_{2}.
For two resistors in parallel: 1/R_{eff} = 1/R_{1} + 1/R_{2}.
Each resistor has resistance R, the equivalent resistance is R.
Kirchhoff's rules
Problem:
Find the equivalent resistance between the points A and B of the circuit
shown in the figure below.
Solution:
Problem:
In the circuit above,
express the current in the 3R resistor in terms of V and R.
Solution:
- Concepts:
Kirchhoff's rules
- Reasoning:
We can find the currents I_{1}, I_{2}, and I_{3}
using Kirchhoff's rules. The junction rule states that the sum of the
currents entering a junction must equal the sum of the currents leaving that
junction. The loop rule states that the sum of the potential differences
around any closed circuit loop must be zero.
- Details of the calculation:
I_{1} + I_{2} = I_{3}, I_{1} = I_{2},
V – 3RI_{3} - RI_{2} = 0. I_{3}
= 2V/(7R).