3. +-3 points SerCP8 18.P.016 My Notes Ask Your Teacher Note: The currents are n

Question

3. +-3 points SerCP8 18.P.016 My Notes Ask Your Teacher Note: The currents are not necessarily in the direction shown. (Use the following as necessary: R1-14.0 , R2 = 4.0 , and V = 36 V.) RI 240 124 (a) Find the current in each resistor of the figure above by using the rules for resistors in series and parallel 4.0 2 resistor 14.0 resistor 24 resistor (b) Write three independent equations for the three currents using Kirchoff's laws: one with the node rule; a second using the loop rule through the battery, the 4.0 resistor, and the 24.0 resistor, and the third using the loop rule through the 14.0 and 24.0 resistors. (Do this on paper. Your instructor may ask you to turn in this work.) Solve to check the answers found in part (a) Need Help?Read It Talk to a Tutor

Explanation / Answer

R1 and 24 ohm are in parallel and their combination is given as

R' = R1 (24)/(R1 + 24) = (14) (24)/(14 + 24) = 8.84 ohm

R' and R2 are in series and their combination is given as

Req = R' + R2

Req = 8.84 + 4 = 12.84 ohm

using ohm's law

i = current coming from the battery = V/Req = 36/12.84 = 2.8 A

since resistance R2 = 4 ohm is in series with the battery, the current in R2 = 4 ohm is same as the current coming from battery

i2 = i = 2.8 A

V2 = i2 R2 = 2.8 x 4 = 11.2 Volts

V1 = Voltage across R1 = V - V2 = 36 - 11.2 = 24.8 Volts

i1 = current in R1 = V1 /R1 = 24.8 /14 = 1.8 A

i24 = current in 24 ohm = i - i1 = 2.8 - 1.8 = 1 A

b)

using node rule

i2 = i1 + i24 eq-1

using loop rule

i2 R2 + i24 (24) = V

4 i2 + 24 i24 = 36

i2 + 6  i24 = 9

i24 = (9 - i2 )/6 eq-2

for third loop

i2 R2 + i1 R1 = V

4 i2 + 14 i1 = 36

i1 = (36 - 4 i2 )/14 eq-3

using eq-1 , eq-2 and eq-3

i2 = (36 - 4 i2 )/14  + (9 - i2 )/6

i2 = 2.8 A

using eq-3

i1 = (36 - 4 i2 )/14  

i1 = (36 - 4 (2.8) )/14  

i1 = 1.8 A

using eq-2

i24 = (9 - (2.8) )/6

i24 = 1 A

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