question_answer

Radius of First orbit of hydrogen is \[0.53\,\overset{o}{\mathop{A}}\,\]. The radius in fourth orbit is :

A)  \[8.48\,\overset{o}{\mathop{A}}\,\]        

B)  \[2.12\,\overset{o}{\mathop{A}}\,\]

C)  \[4.24\,\overset{o}{\mathop{A}}\,\]       

D)  \[0.193\,\overset{o}{\mathop{A}}\,\]

Correct Answer: A

Solution :

Key Idea: In steady state there is no current in capacitor branch. In the given circuit \[2\,\Omega \] and \[3\,\Omega \] resistors are connected in parallel, hence their effective resistance is \[\frac{1}{R}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\] \[\Rightarrow \] \[R=\frac{6}{5}=1.2\,\Omega \] In steady state, no current flows through capacitor branch. So, R will be in series with \[2.8\,\Omega \] resistor. \[R\,=1.2\,\Omega +2.8\,\Omega =4\,\Omega \] From Ohms law \[V=IR\] \[\Rightarrow \] \[I=\frac{6}{4}=1.5\,A\] Potential difference across \[2.8\,\Omega \]. is \[V=2.8\times 1.5=4.2\ \Omega \] Potential difference across a combination of\[2\ \Omega \] and \[3\ \Omega \] is \[=6-4.2=1.8\,V\] Current in \[2\,\Omega ,\,{{i}_{1}}=\frac{1.8}{2}=0.9\,A\] Note: Potential difference across resistors connected in parallel is same.