×

# 3 Resistance Wheatstone Bridge

Consider the Wheatstone Bridge circuit. If the resistances R1 and R2 were zero(i.e if they were not present without altering the connections) then, the ratio of resistances will be 0:0. Will the current passing through G still be zero?

Note by Rajath Krishna R
4 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

If $$R_1 = R_2 = 0 \Omega$$, then $$CAD$$ is a short. So, $$R_k$$, $$R_x$$, and $$G$$ are shorted out of the circuit.

This means that the currents through $$R_k$$, $$R_x$$, and $$G$$ are all $$0 A$$.

- 4 years, 7 months ago

No current will pass through points CD, as the equivalent resistance between those points would be $R_{eq}=\frac{0[R_k+R_x]}{0+ R_k+R_x} = 0$. And since resistance is zero, no current will flow through $$CD$$.

- 4 years, 7 months ago

YES

- 2 years, 11 months ago

yes?

- 2 years, 11 months ago

0

- 3 years, 4 months ago

zero A

- 4 years, 7 months ago