Three resistances of 4,Omega each are connect | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) When three resistors of $4\,\Omega$ are connected in an equilateral triangle,let the corners be $A$,$B$,and $C$.To find the effective resistance b

Vedclass · Accepted Answer

$\frac{8}{3}\,\Omega$

Vedclass · Answer

(D) When three resistors of $4\,\Omega$ are connected in an equilateral triangle,let the corners be $A$,$B$,and $C$.To find the effective resistance between any two corners (e.g.,$A$ and $B$),we observe the circuit configuration.The resistor connected directly between $A$ and $B$ is in parallel with the series combination of the other two resistors (connected between $A-C$ and $C-B$).Resistance of the series branch = $4\,\Omega + 4\,\Omega = 8\,\Omega$.Now,this $8\,\Omega$ branch is in parallel with the $4\,\Omega$ resistor connected directly across $A$ and $B$.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} = \frac{2+1}{8} = \frac{3}{8}$.Therefore,$R_{eq} = \frac{8}{3}\,\Omega$.

Three resistances of $4\,\Omega$ each are connected in the form of an equilateral triangle. The effective resistance between any two corners is:

Similar Questions

The resistance of the series combination of two resistances is $S$. When they are joined in parallel,the total resistance is $P$. If $S = nP$,then the minimum possible value of $n$ is

Different combinations of $3$ resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is:

Four resistances of $100 \ \Omega$ each are connected in the form of a square. Then,the effective resistance along the diagonal points is ............. $\Omega$.

$A$ wire of resistance $R$ is bent to form a circular ring. Find the equivalent resistance between two points $X$ and $Y$ on its circumference. (The angle $XOY = \alpha$ is given.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module