Skip to main content
Electrical Engineering

Return to Question

Formatted equations and neatened up spelling/ grammar
Source Link
edit approved Sep 22, 2014 at 9:43
MrPhooky
edit approved Sep 22, 2014 at 9:43
MrPhooky
  • 2.2k
  • 8
  • 30
  • 57

simple ohm's law questions Simple Ohm's Law Question

some basic electrical questions

do iDo I need a resistanceresistor if V/I = 0 \$\frac{V}{I}=0\$?

assume imI am working on a simple circuit of an ledLED and a buzzer in series where it wouldthe voltage drop would be 2 and 3 volts  , so R = V/I = (5 - (3 + 2))/I \$R=\frac{V}{I}=\frac{(5 - (3 + 2))}{I}\$, in that case do iI need a resistance  ?

wouldWould a current flow in the previous case  ? isn't electricityIsn't current flow because ofdue to the voltage difference of maybe 5v5V and 0v0V on 5va 5V battery  ? soSo when iI drop the voltage of 5v then5V, why would current even flow  ?

alsoAlso in the following picture  :

enter image description here

in thatIn this picture, that 98.90A is where current tends to infinity as resistance is too low, but in the first one, how did it calculate that 49.22mA ?

R = (5-2)V/49mA = 61 ohm\$R = \frac{(5-2)}{49mA} = 61 ohms\$, how did circuit wizardCircuit Wizard add 61 ohms, isn't that way too muchhigh for wire resistance  ?

simple ohm's law questions

some basic electrical questions

do i need a resistance if V/I = 0 ?

assume im working on a simple circuit of an led and buzzer in series where it would drop 2 and 3 volts  , so R = V/I = (5 - (3 + 2))/I , in that case do i need a resistance  ?

would a current flow in the previous case  ? isn't electricity flow because of the voltage difference of maybe 5v and 0v on 5v battery  ? so when i drop the voltage of 5v then why would current even flow  ?

also in the following picture  enter image description here

in that picture, that 98.90A is where current tends to infinity as resistance is too low, but in the first one, how did it calculate that 49.22mA ?

R = (5-2)V/49mA = 61 ohm, how did circuit wizard add 61 ohms, isn't that way too much for wire resistance  ?

Simple Ohm's Law Question

Do I need a resistor if \$\frac{V}{I}=0\$?

I am working on a simple circuit of an LED and a buzzer in series where the voltage drop would be 2 and 3 volts, so \$R=\frac{V}{I}=\frac{(5 - (3 + 2))}{I}\$, in that case do I need a resistance?

Would a current flow in the previous case? Isn't current flow due to the voltage difference of maybe 5V and 0V on a 5V battery? So when I drop the voltage of 5V, why would current even flow?

Also in the following picture:

enter image description here

In this picture, that 98.90A is where current tends to infinity as resistance is too low, but in the first one, how did it calculate that 49.22mA ?

\$R = \frac{(5-2)}{49mA} = 61 ohms\$, how did Circuit Wizard add 61 ohms, isn't that way too high for wire resistance?

Source Link
Andrew
Andrew
  • 279
  • 1
  • 9

simple ohm's law questions

some basic electrical questions

do i need a resistance if V/I = 0 ?

assume im working on a simple circuit of an led and buzzer in series where it would drop 2 and 3 volts , so R = V/I = (5 - (3 + 2))/I , in that case do i need a resistance ?

would a current flow in the previous case ? isn't electricity flow because of the voltage difference of maybe 5v and 0v on 5v battery ? so when i drop the voltage of 5v then why would current even flow ?

also in the following picture enter image description here

in that picture, that 98.90A is where current tends to infinity as resistance is too low, but in the first one, how did it calculate that 49.22mA ?

R = (5-2)V/49mA = 61 ohm, how did circuit wizard add 61 ohms, isn't that way too much for wire resistance ?