Consider the circuit shown in the figure(Figure 1) . Suppose the four resistors
ID: 1650033 • Letter: C
Question
Consider the circuit shown in the figure(Figure 1) . Suppose the four resistors in this circuit have the values R1 = 13 , R2 = 6.7 , R3 = 7.0 , and R4 = 11 , and that the emf of the battery is E = 18 V .
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Part A
Find the current through each resistor using the rules for series and parallel resistors.
Express your answers using two significant figures separated by commas.
I1,I2,I3,I4 = A
Part B
Find the current through each resistor using Kirchhoff's rules.
Express your answers using two significant figures separated by commas.
I1,I2,I3,I4 = A
w 2 3 4Explanation / Answer
part A:
R2 and R4 are in series.
net resistance=R2+R4=6.7+11=17.7 ohms
this is in parallel with R3.
net resistance=17.7*7/(17.7+7)=5.0162 ohms
it is in series with R1.
net resistance=13+5.0162=18.0162 ohms
so current in the circuit=emf/total resistance=18/18.0162=0.9991 A
current through R1=0.9991 A
using current division rule,
current through R3=0.9991*17.7/(17.7+7)=0.71595 A
current through R2=current through R4=0.9991*7/(17.7+7)=0.28315 A
so in 2 significant numbers,
current are:
i1=1 A
i2=0.28 A
i3=0.72 A
i4=0.28 A
part B:
using kirchoff's rule:
let current in left mesh be i1 and right mesh be i2.
assume that both currents are in clockwise direction
writing KVL for left mesh:
18-R3*(i1-i2)-R1*i1=0
==>20*i1-7*i2=18...(1)
writing KVL for right mesh:
R3*(i2-i1)+(R2+R4)*i2=0
==>24.7*i2=7*i1
==>i1=3.5286*i2...(2)
using equation 2 in equation 1,
20*3.5286*i2-7*i2=18
==>63.572*i2=18
==>i2=0.28315 A
then i1=0.9991 A
so current in 2 significant digits:
current through R1=1 A
current through R2=0.28 A
current through R3=0.72 A
current through R4=0.28 A