Consider the circuit shown in the figure below. Use the following as necessary:
ID: 1646247 • Letter: C
Question
Consider the circuit shown in the figure below. Use the following as necessary: R1 = 6.00 , R2 = 1.10 , and V = 11.00 V.
(a) Calculate the equivalent resistance of the R1 and 5.00 resistors connected in parallel.
(b) Using the result of part (a), calculate the combined resistance of the R1, 5.00 , and 4.00 resistors.
(c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00 resistor.
(d) Combine the equivalent resistance found in part (c) with the R2 resistor.
(e) Calculate the total current in the circuit.
A
(f) What is the voltage drop across the R2 resistor?
V
(g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00 resistor.
V
(h) Calculate the current in the 3.00 resistor.
A
Explanation / Answer
Remember:
For series combination
Req = R1 + R2 + R3 +...............
for parallel combination
1/Req = 1/R1 + 1/R2 + 1/R3 + ............
for 2 resistors in parallel it will be
Req = R1*R2/(R1+R2)
Using this Information:
A.
6 and 5 ohm resistors are in parallel,
Rp = 6*5/(6 + 5) = 2.73 ohm
B.
Now Rp and 4 ohm are in series, So
Rs = Rp + 4 = 2.73 + 4
Rs = 6.73 ohm
C.
Now Rs and 3 ohm are in parallel, So
Rp1 = 6.73*3/(6.73 + 3) = 2.08 ohm
D.
Now Rp1 and R2 are in series, So
Req = 2.08 + 1.10 = 3.18 ohm
E.
Using ohm's law:
V = i*R
i = V/Req = 11/3.18 = 3.46 Amp.
F.
Current in R2 and Rp1 will be equal to total current, So
i2 = 3.48 Amp.
V2 = i2*R2
V2 = 3.48*1.1 = 3.83 V
G.
Voltage in parallel is remains same in each resistor, So
Vs = Voltage in 3 ohm = Vp1
Vp1 = 11 - 3.83 = 7.17 V
V in 3 ohm = 7.17 ohm
H.
Current in 3 ohm will be
i3 = V3/R3
i3 = 7.17/3 = 2.39 Amp.