| In the circuit given below. |
 |
| Potential difference of points A and B at time t is |
A) \[\frac{E}{3}\left( 1+\frac{1}{2}{{e}^{-\,t\left( \frac{3R}{2L} \right)}} \right)\]
B) \[\frac{E}{3}\left( 1-\frac{1}{2}{{e}^{-\,t\left( \frac{3R}{2L} \right)}} \right)\]
C) \[\frac{E}{2}\left( 1+{{e}^{-\,t\left( \frac{3R}{2L} \right)}} \right)\]
D) \[\frac{E}{2}\left( 1-{{e}^{-\,t\left( \frac{3R}{2L} \right)}} \right)\]
Correct Answer: A
Solution :
| Effective emf of circuit is \[{{E}_{eff}}=\frac{E}{2}\] |
| Effective resistance of circuit is \[{{R}_{eff}}=\frac{3}{2}R\] |
| So, time constant of circuit is \[\tau =\frac{{{L}_{eff}}}{{{R}_{eff}}}=\frac{2L}{3R}\] |
| Current through inductor at time t is \[i=\frac{E}{3R}\left( 1-{{e}^{-\,t\left( \frac{3R}{2L} \right)}} \right)\] |
| \[iR=\frac{E}{3}(1-{{e}^{-t/a}})\] \[[\because iR=V]\] |
| \[V=\frac{E}{3}(1-{{e}^{t/a}})\] |
| Potential drop across inductor is \[{{V}_{L}}=L\frac{di}{dt}=\frac{E}{2}{{e}^{-t\left( \frac{3R}{2L} \right)}}\] |
| So, \[{{V}_{AB}}={{V}_{AC}}+{{V}_{L}}\] \[=\frac{E}{3}(1-{{e}^{-\,t/\tau }})+\frac{E}{2}{{e}^{-\,t/\tau }}\] |
| \[=\frac{E}{3}\left( 1+\frac{1}{2}{{e}^{-t\left( \frac{3R}{2L} \right)}} \right)\] |
You need to login to perform this action.
You will be redirected in
3 sec
