Download Lesson 8.02 example problems.

Circuits
1. In the figure below, the battery voltage is 12 Volts. Calculate
a. the total equivalent resistance in the circuit
b. the total power dissipated by the entire circuit
10Ω
12V
5Ω
2Ω
3Ω
1Ω
4Ω
15Ω
a. The 5 and 2 Ω resistors are in parallel, forming an equivalent resistance of
1 1 1 7
or R1 = 10 7 = 1.43Ω . This resistor is in series with the 15Ω resistor, forming
= + =
R1 5 2 10
an equivalent resistance of R1,eff = 16.4W. The 1 and 3Ω resistors are also in parallel, forming an
equivalent resistance of
1 1 1 4
= + = or R2 = 3 4 = .75Ω . This resistor is in series with the
R2 1 3 3
10Ω and 4Ω resistor, forming an equivalent resistance of R2,eff = 10 + 4 + 0.75 = 14.75W. This
leaves R1,eff and R2,eff in parallel, forming a final equivalent resistance of
1
1
1
=
+
,
Req 16.3 14.75
which gives Reff = 7.77Ω.
b. Solving for the total current through the circuit, we have VB = ItotReff or Itot = 12V/7.77W = 1.54A.
This means that the battery is delivering power P = ItotV = 1.54A⋅12V = 18.5W. By energy
conservation, this is the total power dissipated by all the resistors.
2. a. Find the equivalent capacitance between points a and b for a group of capacitors connected
as shown in the figure below if C1 = 2µF, C2 = 4 µF, and C3 = 5 µF.
b. If the potential difference between points a and b is 16V, what charge is stored on C3?
a
C1
C1
C3
C2
C2
C2
C2
b
a. Capacitors C1 and C2 on each side of the upper “rectangle” are in series, with an equivalent
capacitance of
1
1
1
1 1 3
=
+
= + = ,so Ceff′ = 4/3 μF. This capacitor is in parallel with a
′
C eq
C1 C 2 2 4 4
Ceq′ on the other side of the “rectangle” and with C3, with a total equivalent capacitance of Ceq″ =
2⋅4/3 + 5 = 23/3 μF. The bottom two C2 capacitors are in parallel with each other, forming an
equivalent capacitance of 8 μF, and this capacitor is in series with Ceq″ = 23/3 μF, forming a final
equivalent capacitance of
1
1
1 3 1 47
, which gives Ceff = 184/43 = 4.27 μF.
=
+ =
+ =
″
C eq C
8 23 8 184
eq
b. If V = 16volts, then the charge on Ceq is given by Q = CeqV = 4.27x16 = 68.5 μC.
This is the same charge on the equivalent capacitance Ceq″ = 23/3 µF, which means the voltage
drop across the upper “rectangle” is given by V = Q/C = 68.5/(23/3) = 8.93 volts.
Since C3 is in parallel with the 2 caps labeled Ceq′, then the voltage across C3 is 8.93volts, and so
the charge on C3 is given by Q3 = C3V = 5x8.93 μC = 44.65µC.
3. Find the current in the following circuit.
a
g
b
R1
c
R
2
R
3
ε1
ε2
R
5
f
R
4
d
e
In this circuit, we have two sources of emf and three resistors. We cannot predict the direction of
the current unless we know which emf is greater. We assume any direction and solve the
problem with that assumption. If the assumption is incorrect, we shall get a negative number for
the current, indicating that its direction is opposite to that assumed.
Let us assume I to be clockwise. The potential drops and increases as we traverse the
circuit in the assumed direction of the current, starting at point a, are as follows (a negative sign
indicates a potential drop):
Vab = -iR2
Vbc = -iR3
Vcd = -ε2
Vde = -iR4
Vef = -iR5
Vfg = ε1
Vga = -iR1.
We encounter a potential drop traversing one of the emfs and an increase traversing the other.
The loop theorem gives ε1 - ε2 = I(R1 + R2 + R3 + R4 + R5), which, when solved for I, yields
I=
ε1 − ε 2
R1 + R2 + R3 + R4 + R5
Notice that if ε2 is greater that ε1, we get a negative number for the current i, indicating that we
have chosen the wrong direction for i. For ε2 greater than ε1, the current is in the counterclockwise
direction. Also notice that while the stronger emf source is discharging, the weaker emf source is
charging.
4. A 6-V battery of negligible internal resistance is used to charge a 2 µF capacitor through a 100
Ω resistor. Find
a. the initial current
b. the final charge
c. the time required to obtain 90 percent of the final charge
a. The initial current is i0 =
ε
R
=
(6 V)
= 0. 06 A
(100 Ω)
b. The final charge is qf = εC = (6 V)(2 µF) = 12 µC.
c. The time constant for this circuit is RC = (100 Ω)(2 µF) = 200 µsec. Setting q = 0.9 εC and
using
q (t ) = q0 (1 − e
−t
RC
) , we obtain
0. 9 εC = εC (1 − e − t / RC )
0. 9 = 1 − e − t / RC
0.1 = e − t / RC
ln( 0.1) = ln( e − t / RC )
− ln(10 ) = − t / RC
t = RC ln(10 ) = ( 200 μs)(2.3) = 460 μs