- #1

Moon_tm

- 5

- 0

## Homework Statement

A light bulb (L1) and a cooker (R1) are connected to the same wire. The wire consists

of two copper conductors with a cross section of 1.5 mm2 and a length of 22.3 m.

When SW1 is closed the heating element of the cooker will be turned on. It has a

power dissipation of 2300 W when it is supplied with 230 V. The light bulb has a

power dissipation of 23 W when it is supplied with 230 V.

As shown on the diagram the wire will from the beginning be supplied with 230 V.

a) What is the voltage drop across light bulb L1 when the cooker is turned off, ie. the

switch SW1 is open?

b) What is the voltage drop across the light bulb L1 when the cooker is turned on, ie.

the switch SW1 is closed?

c) What is the power dissipated by light bulb L1 when the cooker is turned off?

d) What is the power dissipated by light bulb L1 when the cooker is turned on?

Note:

The light bulb and the heating element can be considered to be constant ohmic

resistances.

Specific resistance of copper

## The Attempt at a Solution

+ on paper P=I^2*R and I=V/R --> R=V^2/P

Assumptions:

- Voltage in parallel doesn't change
- Current coming through the wire splits into the branches

I can find out the resistance of the wire, but not the p.d. it produces, since I don't know the current (which is dependent upon how much Voltage is left after the first wire has been passed). Not even voltage division works, since the Resistance of the two appliances is dependent upon the current passing through them and that current is once again, dependent on just how much Voltage remains after the first wire...

Learning to use mathematica for physics:

Problem would produce recursion, since variables were interdependent - R_L and R_R on DV_W, which is dependent on R_Total (which is, wait, dependent on R values of the appliances, which depend on the current, which depends on the voltage left after the first wire...I'm lost, (in a recursive loop, just like tho Mathematica), please help. Need another perspective...

Edit: Now that I'm cooking lunch and thinking about it, V isn't limited to the given value V_rms, (V_max*sqrt(2)/2), so the the appliances could draw the P/V current...

Last edited: