0
\$\begingroup\$

I have some basic question about current. I have this picture here. enter image description here

Why does the ground have more voltage drop across than the cabels connecting the metal poles together? I could not wrap my head around this. All I could assume is that current stays constant, therefore, the voltage drop must increase when there is higher resistance to maintain the same current throughout the circuit. Is this right? If so, I still don't get why. Why does it have to maintain the same current? I thought current was the dependent variable in simple circuit.

\$\endgroup\$
  • \$\begingroup\$ You have some current. And then you have the cable with low resistance and the earth with high resistance. According to Ohm's law the higher resistance will have higher voltage. \$\endgroup\$ – Eugene Sh. Mar 17 '16 at 21:50
  • \$\begingroup\$ But according to the same law, current can change instead of voltage. Why is that? \$\endgroup\$ – steven Mar 17 '16 at 21:56
  • \$\begingroup\$ As I said, regardless of current the ratio of the voltages is the same as the ratio of the resistances. \$\endgroup\$ – Eugene Sh. Mar 17 '16 at 21:59
  • \$\begingroup\$ It cant be regardless of current because the current HAS to stay constant in order for the ratio of the voltages and the ratio of the resistances to remain the same. I am asking why it stays the same. \$\endgroup\$ – steven Mar 17 '16 at 22:01
  • \$\begingroup\$ According to math and Ohms law \$V_1=I\cdot R_1\$, \$V_2=I\cdot R_2\$. Therefore \$\frac{V_1}{V_2}=\frac{I\cdot R_1}{I\cdot R_2}= \frac{R_1}{R_2}\$. Current goes away \$\endgroup\$ – Eugene Sh. Mar 17 '16 at 22:04
1
\$\begingroup\$

The current in the circuit loop is indeed constant. It must be because no electrons are lost and current is just the movement of electrons.

Indeed ground is not a good conductor, actually it is a bad conductor. This results in more voltage drop for the same current compared to a more conductive (less resistance) wire. This is described by Ohm's Law.

The current indeed depends on the circuit. The current is determined by the voltage V and the total resistance R. Then Ohm's Law states: $$I = V / R$$ where I is the current. So a smaller R (less resistance) results in more current.

\$\endgroup\$
  • \$\begingroup\$ Is there a case where electrons are lost? I thought they are never lost. And why does the current HAVE to stay the same? \$\endgroup\$ – steven Mar 17 '16 at 21:58
  • \$\begingroup\$ No electrons cannot get "lost". You can electrically charge an object (static electricity) but then no current flows. In a complicated circuit with many loops it might appear that a current gets lost but it doesn't loops can interact but the total sum is always the same as no electrons are lost. \$\endgroup\$ – Bimpelrekkie Mar 17 '16 at 22:01
  • \$\begingroup\$ I suggest you read a textbook about the basics of electricity, it will all be explained there. \$\endgroup\$ – Bimpelrekkie Mar 17 '16 at 22:21
  • \$\begingroup\$ Kirchoff's Current Law states that the current will be the same at all points in a simple series circuit. The magnitude of that current will vary, depending on the total resistance in the circuit, and on the total applied voltage. \$\endgroup\$ – Peter Bennett Mar 17 '16 at 22:24
0
\$\begingroup\$

A few basics:

  • The source voltage remains fairly constant, 2400 V, in this scenario. The electricity grid is designed on the basis that voltage is constant +/-10% throughout and that guarantees reasonably steady voltage in your home, etc.
  • The current varies depending on the load. This stands to reason: a small lamp will draw a small current because its resistance (to current) is high. A big heater will draw a large current because its resistance is low.

Why does the ground have more voltage drop across than the cables connecting the metal poles together?

  • The answer is that, if the voltages marked on the drawing are correct, the ground resistance must be higher. Let's guess that the cable resistance from the source to the break is 1 Ω. Since the voltage drop across the ground is 239 times the 10 V drop along the wire then the ground circuit must be 239 Ω. This will be made up of the resistance of the contact of the wire with the ground, the ground resistance itself and the earth connection at the source.
  • The cables don't connect the poles together. The cables are insulated from the poles which are there only for support.

All I could assume is that current stays constant, therefore, the voltage drop must increase when there is higher resistance to maintain the same current throughout the circuit. Is this right?

  • No, the voltage remains constant. Current varies with the resistance / load.

I thought current was the dependent variable in simple circuit.

  • Yes, this is correct. The current varies with the load.
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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