Mastering Physics Solutions: Throw the Switch

Mastering Physics Solutions: Throw the Switch

On May 5, 2012, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Throw the Switch

Part A = The potential difference across B drops to zero, The potential difference across A increases by 50%.
Part B = The potential difference across A increases.

Solutions Below:

In this problem ε denotes the emf provided by the source, and R is the resistance of each bulb.

Part A

Bulbs A, B, and C in the figure (Part A figure) are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference?

  • The potential difference across A is unchanged.
  • The potential difference across B drops to zero.
  • The potential difference across A increases by 50%.
  • The potential difference across B drops by 50%.

Two of the above are correct:

The potential difference across B drops to zero, The potential difference across A increases by 50%.

Part B

One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown in the figure. (Part B figure) Bulbs A, B, C, and D are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference?

  • The potential difference across A increases.
  • The potential difference across B doubles.
  • The potential difference across B drops to zero.
  • The potential difference across D is unchanged.

The potential difference across A increases.

Tagged with:  

4 Responses to Mastering Physics Solutions: Throw the Switch

  1. Sara says:

    I do not understand part A. Why does the potential difference across A decrease? Does the switch cut off the flow for B as well? What does the switch do in this situation?

    • Mastering Physics Solutions says:

      Yes – the switch cuts off the current to both the B and C resistors. The potential difference across A must therefore increase. Think about it this way – with the switch open, electricity travels through all three resistors. Assume that each resistor has a resistance of 1 ohm. B and C are in parallel, so their overall resistance is 0.5. And A is in series with them, so the total resistance with the switch open is 1.5. V = IR, so if you have a 6 volt battery, 6 = I * 1.5 and I (current) = 4 amps. Across resistor A, V = IR, so V = 4 * 1. So the voltage drop across A is 4 volts.

      If you close the switch, then there’s a path through the circuit where electricity doesn’t have to go through resistors B or C. Electricity takes the path of least resistance, so B and C are essentially eliminated. We’re assuming that the wire has no resistance. In reality, the wire will have a little bit of resistance, so resistors B and C will have a very little bit of current going through them. But if you assume for sake of argument that the wire has no resistance, then all the electricity will bypass B and C. That means the voltage drop across A has to be 6 volts, an increase of 50%. Hope that helps!

  2. hope says:

    why is in section B: potential across A would increase?

    • Mastering Physics Solutions says:

      Closing the switch bypasses bulb C. So the voltage across A has to drop more, since that source of resistance has been removed.

Leave a Reply

Your email address will not be published.


*

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>