February Break Assignment - Clue #1

Clue #1

Solutions will be posted tomorrow. Here are the solutions to Friday's work:

Problem 3.
a) This is a standard problem where you have to find Req. The trick here is to realize that the 3 ohm and 6 ohm resistors are in parallel since their ends are connected directly together. You should obtain that Req = 6 ohms for the entire circuit. The battery current is therefore 1 ampere, and since the 4 ohm resistor is in series with the battery, it also has a current of 1 ampere.

b) To show that the junction rule holds, you must demonstrate that the current into the junction (which is the current through the 3 ohm and 6 ohm resistors) is equal to the current out (the current going through the 4 ohm resistor). This emans the only thing missing is the current through the 3 ohm and 6 ohm resistors.

You can use Ohm's law to find that the voltage drop across the 4 ohm resistor is (4 ohm)*(1 A) = 4 V, which means the voltage left to drop across the 3 ohm and 6 in parallel is 2 V. Since both resistors are in parallel, they BOTH must have 2 volts across them. This allows you to calculate the current through them by Ohm's law, 2/3 A and 1/3 A respectively. Since 2/3 A + 1/3 A (current in) = 1 A (current out) we show that the node rule holds.

c) The only thing you must show is that for the large loop, the sum of potential differences adds to zero. Going clockwise from the negative terminal of the battery:

6 V - (2/3 A)(3 ohms) - (1 A)(4 ohms) = 0.

The rule holds!


1989B3.

a)
i. 40 W (direct application of P = IV)
ii. 20 W (calculation of power for a constant force, P = F*v)
iii. 60 W (the battery is supplying ALL of the power to the circuit, so it must be equal to the sum of the power used by the other circuit elements.)

b)
i. 20 V (Ohm's law)
ii. 10 V (You know the power is 20 W, and the current through the motor is 2 A. Using P = IV, the voltage must be 10 V)
iii. 30 V (Similar to part (a), the battery is the source of all potential difference in the circuit, so it must be the sum of the potential differences for the motor and resistor.)


c) 15 V (You can get this by calculating the new power required to lift the mass and using P = IV, or by using the fact that it says the voltage is directly proportional to the speed of the motor. )

d) 7.5 ohms (15 V potential difference across the motor, and the battery potential difference is still 30 V. Thus the resistor has 15 V across it, with the same 2 A current. This makes the resistance 7.5 ohms by Ohm's law.)

1983B3

a) 5 ohms
b) i. 4/3 A ii. 2/3 A
c) At point B: 10 V, at point C: -10 V, at point D: -2V
d) 40 W

1982B4

a) clock in parallel with the battery, radio in series with a resistor, and together in parallel with the battery.
b) 600 ohms
c) P = .45 W, energy = 27 J for a minute

Also, here are a couple more circuit problems to make sure you're where you need to be. The answers are included. Click to see them full size (unless your eyes are really THAT good....)