The effective resistance across AB in the net | vedclass

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

Question

(B) $1$. The two resistors of $2 \,\Omega$ each are connected in parallel. Their equivalent resistance $R_1$ is given by: $\frac{1}{R_1} = \frac{1}{2}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) $1$. The two resistors of $2 \,\Omega$ each are connected in parallel. Their equivalent resistance $R_1$ is given by: $\frac{1}{R_1} = \frac{1}{2} + \frac{1}{2} = 1 \implies R_1 = 1 \,\Omega$.$2$. The two resistors of $4 \,\Omega$ each are connected in parallel. Their equivalent resistance $R_2$ is given by: $\frac{1}{R_2} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies R_2 = 2 \,\Omega$.$3$. Now,$R_1$ and $R_2$ are in series. The total effective resistance $R_{net} = R_1 + R_2 = 1 \,\Omega + 2 \,\Omega = 3 \,\Omega$.

The effective resistance across $AB$ in the network shown is .......... $\Omega$.

Similar Questions

$A$ $100\, W$ bulb $B_1$ and two $60\, W$ bulbs $B_2$ and $B_3$ are connected to a $220\, V$ source,as shown in the figure. If $P_1, P_2$,and $P_3$ are the output powers of the bulbs $B_1, B_2$,and $B_3$ respectively,then:

Five identical electric filament lamps numbered $1$ to $5$ are joined in parallel across an ideal source as shown in the figure. When all the bulbs are working, the reading in the ammeter is $3 \,A$. When lamp '$1$' is switched off, the reading of the ammeter is (in $\,A$)

How can resistors be connected? Explain.

As shown in the figure,$13$ resistors each of resistance $R$ are connected. The effective resistance between points $A$ and $B$ is:

$A$ straight wire of resistance $18 \Omega$ is bent in the form of an equilateral triangular loop. The effective resistance between any two vertices of the triangle is (in $Omega$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module