0
$\begingroup$

From the following graph, determine whether (h) will be on or off: graph

I think (h) should be off because there is no potential difference across it. Is this correct?

Edit:

graph 2

Using the method provided by sweber, I should have (d) be on, yet the answer says it is off because there is no current flowing across (d).

Is the answer wrong or am I not getting the method right?

$\endgroup$
3
  • $\begingroup$ What are the circles with X's in them? That's the typical symbol for a mixer. Are those supposed to be mixers? Note also that homework-like questions require than you show what you have tried to solve the problem before the question can be considered valid for this site. Those are the rules we have to prevent a flood of people trying to get us to do their homework. $\endgroup$
    – DanielSank
    Commented Jul 26, 2015 at 17:27
  • $\begingroup$ No, it is an absolute standard symbol for a bulb in electronics. A similar symbol exists for a mixer, but it has three terminals with arrows, does not make sense here, and is used more for sketches of signal flow, not the actual electronic schematic. $\endgroup$
    – sweber
    Commented Jul 27, 2015 at 6:34
  • $\begingroup$ This symbol represents bulb. No this is not for my homework. I have no homework indeed. $\endgroup$
    – Rescy_
    Commented Jul 27, 2015 at 9:54

2 Answers 2

7
$\begingroup$

There IS a potential, and all three bulbs will be on.

First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V.

If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V:

enter image description here

So, bulb 1 and 2 will both encounter a voltage of 5V, while bulb 3 encounters 10V. If the bulbs are identical (and are all made for 10V), bulbs 1 and 2 will be much dimmer than bulb 3.

Another fact: Bulb 1 and 2 act as voltage divider, so even if you remove the vertical blue wire, both horizontal wires will still have a potential of 5V. As result, there will be no current flow through the vertical blue wire. May be, this causes your confusion.

Finally, your circuit is equivalent to this one:

enter image description here

Again, you can remove the horizontal blue wire, as there will flow no current through it.

$\endgroup$
7
  • $\begingroup$ Nice use of colors. $\endgroup$
    – user5402
    Commented Jul 26, 2015 at 12:21
  • $\begingroup$ How did you know that a circle with a cross in it is a bulb? $\endgroup$
    – DanielSank
    Commented Jul 26, 2015 at 22:41
  • $\begingroup$ @DanielSank: because it's the standard symbol for a bulb, just as the long, thin and short, thick lines stand for a battery. It's only strange that the terminals were arranged in a 90°-angle, not straight, at it's usual. $\endgroup$
    – sweber
    Commented Jul 27, 2015 at 6:26
  • $\begingroup$ @sweber This is funny: in eight years of research I guess I've never used a light bulb in a circuit diagram :) In RF design the circle with a cross is a mixer. $\endgroup$
    – DanielSank
    Commented Jul 27, 2015 at 6:44
  • $\begingroup$ May be. However, I have never considered static DC circuits as RF... $\endgroup$
    – sweber
    Commented Jul 27, 2015 at 6:55
1
$\begingroup$

Bulb H is connected to second battery directly. As long as this bulb is not short circuited, there will be some applied potential difference across its terminals and hence it will be ON

$\endgroup$
1
  • $\begingroup$ As long as resistance of bulb 3 is not zero - otherwise the two voltage sources end up back to back (with a "short") $\endgroup$
    – Floris
    Commented Jul 26, 2015 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.