Physics: Kirchoff's Laws?

Question

1. There are 3 resisitors: R1 (103mA) is attached to a circuit that flows to the right into a node which splits into 2 additional resistors (which are in parallel). R2 (77.5mA) is on the top branch. R3, has an unknown current, which I believe is 22.5mA (If I am using the Kirchoff's Law appropriately - if not please let me know).

2. If the resistance on R3 =4.70kOhms, what is the voltage drop across R2 using Kirchoff's Voltage Law?

3. What is the resistance of R2?

4. If the EMF applied to the circuit is 225V, what is the voltage drop across R1?

5. What is the resistance of R1?

I WILL PICK A BEST ANSWER!!!

Thanks in advance :-)

ANY help on ANY of the questions would be greatly appreciated.

[This is from an ALEGEBRA BASED college level physics class for non-physics majors. ]

[I entered them as one question to save points...]

Answer 1

Since R2 and R3 are in parallel they will have the same voltage drop across their node points which I shall call as P and Q
Hence Vpq = I3R3 =25.5mAx4.7kohm = 120V (approx.)
This pd appears across R2 also.
The rest of the p.d. appears across R1 = 225V - 120V = 105V
Since we now know the current and voltage across R1, we can calculate its resistance value.
R1 = 105V/103mA = 1019ohm or approx 1kohm
R2 = 120/77.5mA = 1548ohm 0r 1.55kohm
* Observe that I have used simple Ohm's Law relationship and havent used Kirchoff's laws at all.

* I think you have incorrectly stated the volatge across R1 is 103V. The math just doesn't add up! It should be 105V

Answer 2

Okay, first I think that you have made an error in your subtraction. 103 mA - 77.5 mA = 25.5 mA. That is the current that must flow inthe other praallel resistor, since by KCL, the currents out of the node (the junction of R1, R2 and R3) must equal the currents into the node. Current doesn't just vanish someplace!

Second, for the voltage across R3, use Ohm's Law:

V = IR, R = V/I, I = V/R.

The same voltage appears across R2 since R3 and R2 are in parallel. Use Ohm's Law to find that, also.

Subtract the voltage across R2//R3 from 225, and the remaining voltage appear across R1 (due to KVL, since the sum of all the voltages (drops and rises) must be zero.

225 = (Voltage across R1) + (Voltage across R2//R3)

Finally, use Ohm's law to find the value of R1:

(Voltage across R1) / (current through R1) = resistance of R1.

Good luck!

Answer 3

define Ia because of the fact the present interior the left hand loop going clockwise. define Ib because of the fact the present interior the final hand loop going CCW. the present interior the 4 ohm resistor is Ia+Ib. Now write the loop equations. 12=Ia(4+7)+Ib(4) 9=Ia(4)+Ib(4+8) clean up for Ia and Ib.

Answer 4

Your need a consultation of specialist in theoretic foundations of electro-technics.

Answer 5

TMQ[too many questions]