Question

Can anyone help me with this?
Can anyone find all resistor voltage drops, power consumed, and currents for this circuit? Please help..
will attach circuit

Answer 1

When you come back online and I'm around, I will walk you though it.

Answer 2

finally i see a question on electronics ...

Answer 3

You can do this a couple of ways. One would be to keep simplifying the circuit. You've combined the two series resistors in each branch into one, now combine the two parallel branches into one with a resistor of value \[R = \frac{R_{2+3}R_{4+5}}{R_{2+3}+R_{4+5}}\Omega = \frac{2K*2K}{2K+2K}\Omega = 1K\Omega\]
That gives us the attached circuit which is just a series circuit with two equal resistors.

Answer 4

We use Ohm's law \(V=IR\) to find the current \(I_1\):
\[I_1=V/(R_1+R) = 12 V/(1K\Omega + 1K\Omega) = 6 mA\]

Now we know that the two parallel branches carry equal currents as they had equal resistances, so each branch will carry \(I_1/2 = 3 mA\), and series resistors carry the same current, so the current through \(R_2 = I_{R_2} = \frac{1}{2}I_1 = 3mA\) and similarly for \(R_3, R_4, R_5\).

The more general way when you don't have such a simple circuit is to use Ohm's law and Kirchhoff's circuit laws to write the equations of all of the voltages and currents. The key notions for this problem are that the sum of all the currents into and out of any junction must be 0, and the voltage drop across a resistor is found with Ohm's law, as is the current, and the net voltage drop around every loop is 0. With the original diagram, I would write something like
\[12 V - I_1*R_1 - I_2*R_2-I_3*R_3 = 0\]\[12 V - I_1*R_1 - I_4*R_4 - I_5*R_5 = 0\]\[I_2 = I_3\]\[I_4 = I_5\]\[I_1-I_2-I_4 = 0\]then solve the equations. In this case, that boils down to
\[12-1000I_1-2000I_2 = 0\]\[12-1000I_1-2000I_4 = 0\]\[I_1-I_2-I_4=0\]and when the dust settles, you get the same answers...

Answer 5

The key to the solution is recognizing the circuit for what it is. It is a 10 volt source connected across two resistors in parallel.
Ia = 10/2 kOhm = 5 ma
Ic = 10/5 kOhm = 2 ma
Ib = Ia + Ic = 7 ma
All voltage drops are 10 volts.
Power consumed is P=IV or (7 ma) (10) or 70 milli watts.

Answer 6

@radar Um, you must have been looking at a different diagram than I was.

Answer 7

Yes the circuit I was looking was this one, how it appeared on that link.....I don't know!
Sorry about the confusion.