00:01
In this problem on the topic of circuits, we are shown a resistor of resistance 6 oms that is connected to an ideal battery of emf 12 volts by two copper wires.
00:09
Each of these wires has a length of 20 centimeters and a radius of a millimeter.
00:14
And we want to find the potential difference across the resistor, as well as the two sections of wire, and the rate at which energy is lost to thermal energy, firstly in the resistor, and then to each section of the wire.
00:28
Now for each wire, we have the resistance of the wire rw equal to the resistivity of the material times the length of the wire l over the area a.
00:43
And the area a is the area of a circle, which is pi r squared or the cross -sectional area of the wire.
00:51
And so putting in values, this is the resistivity 1 .69 times 10 to the minus 8.
01:01
Om meters times the length of the meter the wire which is 20 centimeters or 0 .2 meters divided by pi times the radius of the wire 0 .001 meters squared which gives the wires resistance to be 0 .0011 oms.
01:29
Now the total resistive load in the battery will call it r total is 2 times the resistance of the wire plus the load r, which is 2 into 0 .0011 oms plus the external resistance 6 oms, which gives a total resistance of 6 .022 oms...