Saved Bookmarks
| 1. |
(a) Find the current I supplied by the battery in the network as shown in steady state. (b) Find the steady state charge on the capacitor. (c) Find the initial current through the battery. |
Answer» Solution :Here we do not have a simple RC circuit as seen in Section 27.7. So, we cannot apply Eqs. 27-43 and 27-44 here. We have studied in Section 27.7 that an uncharged  capacitor will intially behave LIKE short circuit and after a long time behave lika a broken wire. Calcualations (a) Once the capacitor is CHARGED, no current will go through it and the current through the middle branch of the circuit is zero in steady state. The `4 Omega` resistor will have no current in it and may be omitted for current analysis. The `2 Omega and 6 Omega` resistors are therefore connected in series and hence `i=(2V)/(2 Omega+6Omega)=0.25A` (B) The potential drop the across the `6Omega" resistor is "6Omega =0.25A=1.5V`. As there is no current in the `4 Omega` resistor there is no potential drop across it. The potential difference acrossthe capacitor is, therefore, 1.5V. The charge on this capacitor is `Q=CV=2muF xx 1.5V=3muC` (c) At the initial moment, an uncharged capacitor can be assumed to be short circuited. So, the EQUIVALENT reisistance of the circuit will be `4.4Omega`. The current through the battery will be `(2)/(4.4)A=0.45 A` |
|