The equivalent resistance of the arrangement | vedclass

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

Question

(B) $1$. First,simplify the upper branch: The resistors $8\,\Omega$,$16\,\Omega$,and $16\,\Omega$ are in parallel. Their equivalent resistance $R_1$ i

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) $1$. First,simplify the upper branch: The resistors $8\,\Omega$,$16\,\Omega$,and $16\,\Omega$ are in parallel. Their equivalent resistance $R_1$ is given by $\frac{1}{R_1} = \frac{1}{8} + \frac{1}{16} + \frac{1}{16} = \frac{2+1+1}{16} = \frac{4}{16} = \frac{1}{4}$,so $R_1 = 4\,\Omega$.$2$. This $R_1$ is in series with the $20\,\Omega$ resistor,so the total resistance of the upper branch is $R_{upper} = 4\,\Omega + 20\,\Omega = 24\,\Omega$.$3$. Next,simplify the lower branch: The resistors $9\,\Omega$ and $18\,\Omega$ are in parallel. Their equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{9} + \frac{1}{18} = \frac{2+1}{18} = \frac{3}{18} = \frac{1}{6}$,so $R_2 = 6\,\Omega$.$4$. This $R_2$ is in series with the $6\,\Omega$ resistor,so the total resistance of the lower branch is $R_{lower} = 6\,\Omega + 6\,\Omega = 12\,\Omega$.$5$. Finally,the upper branch $(24\,\Omega)$ and lower branch $(12\,\Omega)$ are in parallel between points $A$ and $B$. The equivalent resistance $R_{AB}$ is $\frac{1}{R_{AB}} = \frac{1}{24} + \frac{1}{12} = \frac{1+2}{24} = \frac{3}{24} = \frac{1}{8}$,which gives $R_{AB} = 8\,\Omega$.

The equivalent resistance of the arrangement of resistances shown in the figure between the points $A$ and $B$ is ............... $\Omega$.

Similar Questions

The total current supplied to the circuit as shown in the figure by the $5 \, V$ battery is: (in $, A$)

A current of $12 \, A$ flows into a system of resistors as shown in the figure. The potential difference between $A$ and $C$ is (in $V$)

In the circuit shown,the resistances are given in ohms and the battery is assumed ideal with $emf$ equal to $3.0 \ V$. The resistor that dissipates the most power is

$A$ new flashlight cell of $e.m.f.$ $1.5\, V$ gives a current of $15\, A$,when connected directly to an ammeter of resistance $0.04\,\Omega$. The internal resistance of the cell is ........... $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module