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Homework Help: Problem calculating the current

  1. Jan 21, 2009 #1
    Hello All

    I’m Umair and I am new in this forum. And I am doing my First year in BSc.

    I have a problem solving this problem can any one help me. With solution I am very kindly thanks of him.

    http://ihost.djmusiq.com/pics/9c34adb7675c57ff4fd71bdf203742d7.JPG [Broken]

    1. Calculating the current supplied by battery?

    thanks a lot
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 21, 2009 #2


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    Science Advisor

    Hi Umair.

    The 5 resistors on the righthand side of that diagram form a circuit topology known as a "H Bridge" (or some people prefer to refer to it simply as a bridge).

    Look carefully and you'll see that you can redraw those 5 resistors in the arrangement of the 5 segments of the letter "H" (there is a 500 and a 100 ohm resistor on one vertical side of the "H", a 100 and a 500 ohm on the other vertical side of the "H", and finally the "bridge" element being the 300 ohm resistor forming the horizontal arm of the "H"). Note that this gets rid of the crossed wires and makes the circuit much easier to visualize.

    As you've probably already realised the bridge topology, though simple looking, is not so easy to solve using the elementary techniques of combining resistors in series and parallel to "collapse" the network into a simlper equivalent (in the way you might do with a simple "ladder" topology).

    To solve this circuit you're best bet in general is to write the 3 mesh equations out in full and solve the simulatanious equations. Another option is to take either of the 3 resistor in the "H" that form a "Y" network and use a star-delta transform to convert them to an equivalent "Delta" network. This actually allows the circuit to be collapsed into a single equivalent resistor.

    If both the above techniques are beyond where you're currently at with the course then you could try invoking symmetry to help. There's an anti-symmetry between the L and R sides of this particular bridge that might be able to exploited, (though personally I'd go straight for the mesh equations).
    Last edited: Jan 21, 2009
  4. Jan 30, 2009 #3
    There are a total of 4 nodes in this circuit. You can use Node-Voltage analysis using the "bottom" node as the reference node. You will only have 2 node-voltages to solve for. Write the 2 KCL equations for these nodes (the 300 ohm resistor won't be included in these equations) and solve. Solve for the source current by dividing the voltage across the source resistance by 100 ohms.
  5. Feb 2, 2009 #4
    mplayer, you are correct in stating the 300 ohm resistor isn't used (as it is actually shorted as drawn). However, dividing the 100 volt source voltage by the 100 ohm resistor as you suggested will yield incorrect source current.

    The upper two resistors (500 & 100 ohms) are in parallel with each other just as the lower two resistors (500 & 100 ohm) are in parallel with each other. The two paralleled sets of resistors are also in series with each other, as well as in series with the 100 ohm resistor that's connected to the positive terminal of the battery. After deriving the equivalent parallel resistance for each paralleled set of resistors, add all 3 resistance values together to yield the total series circuit resistance, then divide the 100 volt source voltage by the total series resistance.

    Hint: The total series resistance is less than 300 ohms, but more than 200 ohms.
  6. Feb 3, 2009 #5
    I'm pretty sure what I stated previously was correct, but I think I see where my wording is vague. I didn't say to divide the source voltage by 100 ohms to solve for the source current. I said solve for the source current by dividing the voltage across the source resistance by 100 ohms. I am referring to the resistor connected in series with the voltage source. The voltage that needs to be divided by 100 ohms isn't the source voltage, but the voltage across that source resistance. That voltage can be determined by writing and solving the 2 KCL equations for the unknown node-voltages. Once those values are known, the voltages across every element in the circuit can be easily determined.
  7. Feb 3, 2009 #6
    With all due respect mplayer, I wouldn’t have stated you were incorrect unless it were so.

    You stated: "Solve for the source current by dividing the voltage across the source resistance by 100 ohms."


    1. The circuit has no need to derive or employ the use of “source resistance”. See definition of “source resistance” below.

    2. Additionally, your statement; "Solve for the source current by dividing the voltage across the source resistance by 100 ohms" fails to make any coherent sense.

    Definition of “source resistance”

    “Source resistance”, also referred to as “internal resistance”, is that resistance which is inherent within the device that produces the electrical potential such as a battery, a power supply, a generator, an alternator, a solar cell, etc...

    This type of circuit is commonly presented to students to see if they can eliminate redundant components (i.e. the 300 ohm resistor) and resolve the “total series resistance” of the circuit. “If/when” the student realizes the nature of the redundant 300 ohm resistor; the student should then realize there are just two sets of identical paralleled resistors in series with each other as well as in series with the 100 ohm resistor. Next, the student should calculate the equivalent resistance of one set of the paralleled resistors:

    Equivalent resistance = 1 / [(1 / R1) + (1 / R2)]

    1 / [(1 / 500 ohms) + (1 / 100 ohms)] = 83.333… ohms

    Since the other set of paralleled resistors utilize the very same two resistor values, the student realizes they can simply use the same equivalent resistance value twice when adding up the series resistance to yield the “total series resistance”:

    83.333 ohms + 83.333 ohms + 100 ohms = 266.666 ohms

    Finally, the 100 volt source voltage is divided by the total series resistance to derive the circuit current:

    100 volts / 266.666 ohms = .375 amperes circuit current
  8. Feb 3, 2009 #7
    Sorry, I was confused about the exact meaning of source resistance then. I assumed that the source resistance was being modeled by that single 100 ohm resistor in series with the voltage source. I thought I heard my professor refer to a resistor connected in that way as 'source resistance', I must have been mistaken or he must have used the term incorrectly.

    Anyway, the voltage across that single resistance (the one connected in series with the voltage source) divided by its resistance (100 ohms) should still yield the current in question, right? That's what I meant by my previous incoherent statement. I just want to make sure I'm understanding the concept properly and conveying understanding semi-effectively (trying to get better thanks to your constructive criticism).

    The following is just pasted from Wikipedia; is it the same as source resistance? This was the concept I was thinking of when I was referring to source resistance in my previous posts.

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