Consider the circuit shown in the figure(Figure 1) . Suppose the four resistors
ID: 1659586 • Letter: C
Question
Consider the circuit shown in the figure(Figure 1) . Suppose the four resistors in this circuit have the values R1 = 12 , R2 = 7.2 , R3 = 7.1 , and R4 = 12 , and that the emf of the battery is E = 15 V .
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.
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
from the circuit R2 and R4are in series
R24 = 7.2 + 12=19.2 ohm
this R24 is parallel with R3
1/R234 = 1/19.2+ 1/7.1
R234 = 5.183 ohm
the total reisistance of the circuit is
R net = R234 + R1 = 5.183 + 12 = 17.183 ohm
Apply Ohm's law
the current in the circuti
I = V/R net = 15/17.183 = 0.872 A
The potential drop across R1 is
V1 = I R1 = 0.872 ( 12) = 10.47 V
across V234 = I R234 = 0.872 ( 5.183) = 4.519 V
the current through R3 is
I3 =V234/R3 = 4.519/7.1 = 0.636 A
since R2 and R4 are in series the current will be same in both resistors
I2 = I4 = I- I3
=0.872-0.636
=0.235 A
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(b)
Apply KCL
I1= I2 + I3
I1 R1 + I3 R3= V
(12) I1 + 7.1 ( I3) = 15 ....(1)
Appy KVL
I1 R1 + I2 R2 + I4 R4 = E
12 (I1) + I2( R2 + R4) = E
12 (I1) + I2 ( 19.2) = 14 ....(2)
equate (1) and (2)
12 (I1) + I2 ( 19.2)=(12) I1 + 7.1 ( I3)
I2 = (7.1) I3/19.2
I1 = (7.1) I3/19.2+I3
12( (7.1) I3/19.2) + I3) + 7.1 (I3) = 15 V
I3 = 0.637 A
I2 = (7.1)(0.637)/19.2
=0.235 A
I4 = 0.235 A
I1=0.235 A+0.637 A=0.872 A