In the circuit in the figure, a battery supplies Vemf = 21.0 V and R1 = 6.00 ?,
ID: 2156928 • Letter: I
Question
In the circuit in the figure, a battery supplies Vemf = 21.0 V and R1 = 6.00 ?, R2 = 5.00 ?, and L = 6.00 H. Figure: http://www.webassign.net/bauerphys1/29-p-046.gifCalculate each of the following a long time after the switch is closed:
a) the magnitude of the current flowing out of the battery?
b) the magnitude of the current through R1
c) the magnitude of the current through R2
d) the magnitude of the potential difference across R1
d) the magnitude of the potential difference across R2
f) the magnitude of the potential difference across L
g) the rate of current change across R1
Explanation / Answer
a) After long time after the switch is closed,the inductor acts as a non-resistant wire, hence there are only two resistances RR and r2 in parallel, hence,the resultant resistance is R=(R1*R2)/(R1+R2), which is here, 6*5/11=30/11= 2.72 ohms.
hence, total current is V/R=21/2.72=7.7 amperes.
b)the magnitude of current through R1 is I*R2/(R1+R2),where I is 7.7 amperes., hence current through R1 is 3.5 amperes.
c)the magnitude of current through R2 is I*R1/(R1+R2),where I is 2.72 amperes., hence current through R2 is 4.2 amperes.
d)Magnitude of potential difference across R1 is V1=R1*I1,where I1 is 3.5 amperes, hence V1=21 volts.
e) Magnitude of potential difference across R2 is V2= V1=21 volts(since,parallel resistors and hence same voltages).
f)since the inductor acts as a short circuit after a long time,potential difference across it is 0 volts.
g)current does not change anymore after a long time, hence rate of current change is 0.
rate me please.