In the adjoining circuit,the potential differ | vedclass

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

Question

(A) The circuit consists of a parallel combination of a branch containing ($6\, \Omega$ and $3\, \Omega$ in parallel,followed by $2\, \Omega$ in serie

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The circuit consists of a parallel combination of a branch containing ($6\, \Omega$ and $3\, \Omega$ in parallel,followed by $2\, \Omega$ in series) and a branch containing $4\, \Omega$.First,calculate the equivalent resistance of the upper branch. The $6\, \Omega$ and $3\, \Omega$ resistors are in parallel,so their equivalent resistance $R_p$ is given by $R_p = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2\, \Omega$.This $2\, \Omega$ is in series with the other $2\, \Omega$ resistor,so the total resistance of the upper branch is $R_{upper} = 2\, \Omega + 2\, \Omega = 4\, \Omega$.The lower branch also has a resistance of $4\, \Omega$.Since both branches have equal resistance $(4\, \Omega)$,the total current of $2\, A$ splits equally between them. Thus,$1\, A$ flows through the upper branch.The potential difference across the parallel combination of $6\, \Omega$ and $3\, \Omega$ is $V = I_{upper} 	imes R_p = 1\, A 	imes 2\, \Omega = 2\, V$.Since the $6\, \Omega$ and $3\, \Omega$ resistors are in parallel,the potential difference across each is the same,which is $2\, V$.

In the adjoining circuit,the potential difference across $3\, \Omega$ is ............... $V$.

Similar Questions

Which of the following elements shows polymorphism?

The following reaction is an example of a ....... reaction. $CH_2Br-CH_2Br + 2KOH(alc.) \xrightarrow{\Delta} CH \equiv CH + 2KBr + 2H_2O$

The incentre of the triangle with vertices $(1, \sqrt{3})$,$(0, 0)$ and $(2, 0)$ is

At $298 \ K$,the solubility product of $PbCl_2$ is $1.0 \times 10^{-6}$. What will be the solubility of $PbCl_2$ in moles/litre?

The rate constant,$k$,of a zero order reaction $2 NH_3(g) \xrightarrow[1130 \ K]{Pt} N_2(g) + 3 H_2(g)$ is $y \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. The rate of formation of hydrogen (in $mol \ L^{-1} \ s^{-1}$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module