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
10
Ω
5
Ω
2
Ω
12V
1
Ω
4
Ω
3
Ω
15
Ω a. The 5 and 2 Ω resistors are in parallel, forming an equivalent resistance of
1
R
1
=
1
5
+
1
2
=
7
10 or R
1
=
10 7
=
1 .
43
Ω
. This resistor is in series with the 15 Ω resistor, forming an equivalent resistance of R
1,eff
= 16.4W. The 1 and 3 Ω resistors are also in parallel, forming an
1 equivalent resistance of
R
2
=
1
1
+
1
3
=
4
3 or R
2
=
3 4
=
.
75
Ω
. This resistor is in series with the
10 Ω and 4 Ω resistor, forming an equivalent resistance of R
2,eff
= 10 + 4 + 0.75 = 14.75W. This leaves R
1,eff
and R
2,eff in parallel, forming a final equivalent resistance of
1
R eq
=
1
16 .
3
+
1
14 .
75
, which gives R eff
= 7.77
Ω . b. Solving for the total current through the circuit, we have V
B
= I tot
R eff
or I tot
= 12V/7.77W = 1.54A.
This means that the battery is delivering power P = I tot
V = 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 C
1
= 2µF, C
2
= 4 µF, and C
3
= 5 µF. b. If the potential difference between points a and b is 16V , what charge is stored on C
3
? a
C
1
C
1
C
3
C
2
C
2
C
2
C
2 b a. Capacitors C
1
and C
2
on each side of the upper “rectangle” are in series, with an equivalent capacitance of
1
C
′ eq
=
1
C
1
+
1
C
2
=
1
2
+
1
4
=
3
4
,so C eff
′ = 4/3 μ F. This capacitor is in parallel with a
C eq
′ on the other side of the “rectangle” and with C
3
, with a total equivalent capacitance of C eq
″ =
2 ⋅ 4/3 + 5 = 23/3 μ F. The bottom two C
2 capacitors are in parallel with each other, forming an equivalent capacitance of 8 μ F, and this capacitor is in series with C eq
″ = 23/3 μ F, forming a final equivalent capacitance of
1
C eq
=
1
C eq
″
+
1
8
=
3
23
+
1
8
=
47
184
, which gives C eff
= 184/43 = 4.27 μ F. b. If V = 16volts, then the charge on C eq
is given by Q = C eq
V = 4.27x16 = 68.5 μ C.
This is the same charge on the equivalent capacitance C eq
″ = 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 C
3
is in parallel with the 2 caps labeled C eq
′ , then the voltage across C
3
is 8.93volts, and so the charge on C
3
is given by Q
3
= C
3
V = 5x8.93 μ C = 44.65µC.

3. Find the current in the following circuit. g a
R
1
R
2 b c
R
3
ε
1 ε
2
R
5
R
4 f 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):
V
V cd
= ε
2
V ab ef
= iR
= iR
5
2
V bc
= iR
3
V de
= iR
V fg
= ε
1
4
V ga
= iR
1
.
We encounter a potential drop traversing one of the emfs and an increase traversing the other.
The loop theorem gives ε
1
- ε
2
= I( R
1
+ R
2
+ R
3
+ R
4
+ R
5
), which, when solved for I, yields
I
=
ε ε
1
−
2
R
1
+
R
2
+
R
3
+
R
4
+
R
5
Notice that if ε
2
is greater that ε
1
, we get a negative number for the current i , indicating that we have chosen the wrong direction for i . For ε
2
greater than ε
1
, 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 i
0
=
R
=
( 6 V)
=
A b. The final charge is q f
= ε
C = (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 ε C and using
( )
q
0
(1
e
− t
RC )
, we obtain
ε
C
= ε
C ( 1
− e
− t RC )
0 9
= − e
−
= e
−
0 1
= ln( e
−
− ln( 10 )
= −
) t
=
RC ln( 10 )
=
( 200
=
460