InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Compare the power used in the 2 Omega resistor in each of the following circuits: (i) a 6V battery in series with 1 Omega and 2 Omega resistors, and (ii) a 4V battery in parallel with 12 Omega and 2Omega resistors. |
|
Answer» Solution :(i) When a `2 Omega` resistors a joined to a 6V battery in series with `1 Omega and 2 Omega` resistors, total resistance of the combination `R_s=2+1+2= 5 Omega` `therefore` CURRENT in the CIRCUIT `I_1=(6V)/( 5 Omega)=1.2 A` `therefore` Power used in the `2 Omega` resistors `P_1=I_1^2R=(1.2)^2 times 2=2.88W` (ii) When `2 Omega` resistors is joined to a 4V battery in parallel with `12 Omega and 2 Omega` resistors, current flowing in `2 Omega` resistor is INDEPENDENT of the other resistors. `therefore` Current flowing through `2 Omega` resistor `I_2=(4V)/(2 Omega)=2A` `therefore` Power used in the `2 Omega` resistor `P_2=I_2^2 R=(2)^2 times 2=8W` `therefore P_1/P_2=(2.88W)/(8W)=0.36:1` |
|