Engineering Simple circuit - Finding Equivalent Resistance

AI Thread Summary
To find Vo in the given circuit with a current of 15 mA and various resistors, the first step is to determine the equivalent resistance (R_eq). The user initially calculates R_eq as 13.8 kΩ by summing resistances in series, but is uncertain about the role of the 2kΩ and 10kΩ resistors due to a gap in the circuit. It is suggested to suppress the current source and analyze the circuit again, focusing on how current divides between branches based on resistance. The potential drop across resistors in series is proportional to their resistance, which can help in calculating the voltage across each branch. A thorough understanding of standard resistance formulas is essential for verifying the total resistance and potential in the circuit.
jmcmillian
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Homework Statement


Find Vo in the circuit.
Image of Circuit Diagram Attached
Variables: Vo, Io
Given: Current (15 mA)
Resistors (300\Omega, 300\Omega, 2k\Omega, 10k\Omega, 200\Omega, 3k\Omega


Homework Equations


Resistors in Series: Ra+Rb+Rc...+Rn
Resistors in Parallel: [(1/Ra)+(1/Rb)+(1/Rc)+...+(1/Rn)]^-1
V_j = I_j*R_j; I_j = R_eq*I

The Attempt at a Solution


I've decided my approach in this problem should be:
(1) Find Equivalent Resistance (R_eq)
(2) Use the formula I_j=R_eq*I to determine I_o
(3) Use I_o to find V_o, as V_o= I_o*R_o

My problem is that I really am unsure how to find the equivalent resistance in this particular problem. What's messing me up is the 2k\Omega and 10k\Omega resistors...and the gap that is between them. I've just never done a circuit like this I guess.

My attempt to find the equivalent is below:
In Series, Left Side:300 and 300 ohm = 600 ohm
In Series, Right Side:200 and 1000 ohm = 1200 ohm
Simplified Resistor: 600 ohm on Left, 1200 ohm on right, 2k and 10k ohm on top.
The equivalent can be found by adding up in series: 600, 1200, 2000, and 10000 = 13,800ohms = 13.8kOhm. Is that correct? If not, how do I attack the equivalence problem?
 

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Hint :-

Suppress the current source and then analyze the circuit again.

Hope this is useful. :smile:
 
It may not correspond to systematic teachings, but the way automatic for me would be to say the 2k and 10k ohm resistors play no part in the problem since no current flows through them. The current merely divides itself flowing through the two branches proportionately to their conductances.

Work out how it divides itself.


There is the same potential drop for the two branches. The potential drop in two resistors in series (in either of the branches) is proportional to the resistance, e.g. it will be equal for the two resistors in the left-hand branch. Work out what it is in each case according to V = IR.

You can work out the total resistance of this circuit from standard formulae you should know and understand, and hence the total potential. Do this as a check, it should come out agreeing with (voltages added up in) previous results. I hope this is not telling you too much according to the rules.
 
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