Figure (a) shows a circuit diagram consisting of three resistor R1=1.0 k?, R2=2.
ID: 1352392 • Letter: F
Question
Figure (a) shows a circuit diagram consisting of three resistor R1=1.0 k?, R2=2.2 k? and R3=4.7 k?, and a real battery with emf E =4.0 V and internal resistance r=100 ?.
a. (6) Find the equivalent resistance R123 of the three external resistors and draw the simplified circuit diagram inside the blank box below.
b. (3) Write Kirchoff’s loop rule for the simplified circuit.
c. (3) Find the total current i through the battery.
d. (3) What is the potential difference Vab between the terminals a and b of the battery?
e. (6) Find the currents , and through the resistors R1, R2 and R3, respectively.
f. (4) What will be the time constant ? of the circuit shown in figure (b) where the battery is replaced by a charged capacitor with capacitance C=1.0 µF ?
RI R, R,Explanation / Answer
R1 and R2 are in series
equivalent resistance=R1+R2=3.2 kilo ohms
it is in parallel with R3
equivalent resistance=R=4.7*3.2/(4.7+3.2)=1.9038 kilo ohms
hence the final circuit will be battery connected with a 1.9038 kilo ohms resistance
(battery means ideal votlage source in series with internal resistance 100 ohms)
part b:
assuming current in the circuit is i A,
kirchoff's voltage law can be written as
E-i*(r+R)=0
==>4-i*(100+1.9038*1000)=0
==>4-2003.8*i=0
part c:
current=4/2003.8=1.9962 mA
part d:
between a and b , potential difference=E-i*r
=4-1.9962*0.001*100=3.80038 volts=3.8 volts(approximately)
part e: using current division rule,
current through the series combination of R1 and R2=1.9962*R3/(R1+R2+R3)
=1.1876 mA
then current through R3=1.9962 mA-1.1876 mA=0.8086 mA
hence current through different resistors are:
R1: 1.1876 mA
R2: 1.1876 mA
R3: 0.8086 mA
part f:
time constant=equivalent resistance*capacitance
=1.9038*1000*1*10^(-6)=1.9038 ms
hence time constant of the circut will be 1.9038 ms
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