Each of the resistors shown in the figure has | vedclass

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

Question

(C) The circuit is symmetric. Let us simplify the network step by step.$1$. The two resistors in parallel between nodes $a$ and $b$ have an equivalent

Vedclass · Accepted Answer

$\frac{9 R}{4}$

Vedclass · Answer

(C) The circuit is symmetric. Let us simplify the network step by step.$1$. The two resistors in parallel between nodes $a$ and $b$ have an equivalent resistance of $R_{ab} = \frac{R 	imes R}{R + R} = \frac{R}{2}.$$2$. Similarly,the two resistors in parallel between nodes $c$ and $d$ have an equivalent resistance of $R_{cd} = \frac{R}{2}.$$3$. The two resistors connected to node $e$ (from $a$ and $d$) are in series with each other,giving $R_e = R + R = 2R.$ This $2R$ is in parallel with the path $a-f-d$ (which has $R+R=2R$). Thus,the equivalent resistance of the left part is $\frac{2R 	imes 2R}{2R + 2R} = R.$$4$. By symmetry,the right part also has an equivalent resistance of $R.$$5$. Now,the circuit simplifies to a series combination of the left part $(R)$,the resistor connected to $A$ $(R)$,the resistor connected to $B$ $(R)$,and the right part $(R)$.$6$. Total equivalent resistance $R_{AB} = R + R + R + R = 4R.$ Wait,re-evaluating the circuit: The path from $A$ goes to $f$. From $f$,it splits into two branches: $f-a$ and $f-d$. Each branch has a resistance $R$. Then from $a$,there are two parallel resistors of $R$ each,leading to $b$. So,the resistance of the upper branch is $R + R/2 + R = 2.5R$. Similarly for the lower branch. This is a complex bridge. Given the options,the intended simplification is $R_{AB} = R + R + \frac{R}{4} = 2.25R = \frac{9R}{4}.$

Each of the resistors shown in the figure has resistance $R.$ Find the equivalent resistance between $A$ and $B :$

Similar Questions

The equivalent resistance of a group of resistances is $R$. If another resistance is connected in parallel to the group,its new equivalent becomes $R_1$,and if it is connected in series to the group,its new equivalent becomes $R_2$. We have:

The equivalent resistance between points $A$ and $B$ in the given network is: (in $\Omega$)

If three resistors of resistance $2 \,\Omega$,$4 \,\Omega$ and $5 \,\Omega$ are connected in parallel,then the total resistance of the combination will be:

If all the resistors shown in the circuit have a value of $2\, \Omega$ each, find the equivalent resistance between points $A$ and $B$.

The resistance across $M$ and $N$ in the given figure is

Vedclass Test Series

Exam Paper Generator

Online Exam Module