**Problem 1**

The battery has a rating of x A-hr. This implies that it can deliver x A for 1 hour. 1 A is per definition equal to a 1.6 10-19 C/s. The total charge that moves through the battery is thus equal to

**Problem 2**

The total charge that flows through the battery crosses a potential difference of 12 V. The associated change in its potential energy is thus equal to

This is the total electrical energy that is delivered by the battery.

**Problem 3**

Assume that the total length of the telephone wire is L, and the short occurs at a distance x from C. The total length of cable between C and D is equal to 2 x. The resistance between C and D is thus equal to:

where r is the resistivity of the wire and A is its cross sectional area. The total length of cable between A and B is equal to 2 (L - x). The resistance between A and B is thus equal to

The ratio of the resistance is equal to

This equation can be used to determine x:

**Problem 4**

Use a coordinate system to describe the cone, which has the left-side of the cone located at x = 0 m. Consider a slice of the cone located a distance x from the origin, with thickness dx and radius r where

The resistance of this slice is equal to

The total resistance of the cone can be obtained by integrating x between x = 0 and x = h:

Note: since each of the slices is in series, the total resistance is just the sum of the resistance of each slice (thus the integral of dR)

**Problem 5**

**NOTE: the following solution is correct for the circuit
shown on the assignment of Frank Wolfs. Different diagrams might
appear on different assignments, but the approach will be the
same.**

The net current at each branch must be zero. This immediately implies that:

or

Now consider the loop with resistors 1 and 2 (and go around this loop in a direction given by the current I2). Kirchhoffs' law states that the net emf around this loop must be equal to zero. Thus:

Do the same thing for the loop with resistors 2 and 3 (and go around this loop in a direction given by the current I2):

We now have two equations with two unknown. The current I1 is equal to

**Problem 6**

Consider the following three current loops, and assume that the current in each loop is flowing in a clockwise direction:

- Loop 1: resistor 1, resistor 5, and resistor 3
- Loop 2: resistor 2, resistor 4, and resistor 5
- Loop 3: emf, resistor 3, and resistor 4

For loop 1 we require that:

This equation can be rewritten as

For loop 2 we require that

This equation can be rewritten as

For loop 3 we require that

A simple relation between the current through resistor 1 and the current through resistor 2 can be obtained by requiring that the potential difference across them is equal to the applied emf:

A second relation between these two currents between these two current can be found by multiplying the equation for loop 1 by -R4 and adding this to the equation for loop 2 multiplied by R5:

Combining the last two equations we can eliminate the current through resistor 2:

The current through resistor 1 is thus equal to

**Problem 7**

The current through resistor 2 is equal to

The current through resistor 5 is thus equal to

Therefore, the voltage across resistor 5 is equal to

**Problem 8**

**NOTE: the following solution is correct for the circuit
shown on the assignment of Frank Wolfs. Different diagrams might
appear on different assignments, but the approach will be the
same.**

Consider the current loop from a to b to c to d to f to g and back to a. The sum of emfs and potential drops around this loop must be equal to 0 (Kirchhoffs' rule). Thus

This equation can be used to determine Vag:

**Problem 9**

**NOTE: the following solution is correct for the circuit
shown on the assignment of Frank Wolfs. Different diagrams might
appear on different assignments, but the approach will be the
same.**

Consider the branch with resistors R2, R3, and R4. The net resistance of R3 and R4 in parallel is equal to:

The net resistance of R34 in series with R2 is equal to

The current through this chain is equal to

A fraction of this current is going through resistor R4. The relation between the current through R3 and the current through R4 is:

The fraction of the total current that goes through R3 is equal to:

Thus

**Problem 10**

Consider two clockwise current loops; loop 1 with resistor R1 and emf source [epsilon]1, and loop 2 with resistor R2 and emf sources [epsilon]1 and [epsilon]2. For loop 1 we obtain according to Kirchhoff:

Thus

**Problem 11**

For loop 2 we obtain according to Kirchhoff:

Thus

**Problem 12**

The current going through emf e1 is equal to

The power delivered by emf e1 is thus equal to