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Thursday, 18 July 2019

Resistors in Series and Parallel Problems

Resistors in series:

A circuit is said to be a series circuit if the current flowing through all the elements is the same.

In figure three resistors R1, R2, R3 are connected in series. So the current (I ) flowing through each resistor is same. But the voltage drop across each resistor is different due to its different resistance value. The total voltage in this circuit is ‘V’

According to Ohm’s Law , V = IR

Voltage drop across R1 is       V1 = IR1
Voltage drop across R2 is       V2 = IR2
Voltage drop across R3 is       V3 = IR3

Total voltage               

V = V1 + V2 + V3
    = IR1 + IR2 + IR3
    = I ( R1 +R2 + R3 )

IR = I ( R1 + R2 + R3 )
R = ( R1 + R2 + R3 )

 The total (or) equivalent resistance is equal to the sum of all the resistances which are connected in series

Resistors in Parallel:

A circuit is said to be a parallel circuit if the potential difference ( voltage ) across all the elements is the same.

In figure three resistors R1, R2, R3 are connected in parallel . So the potential drop across each resistor is same. But the current flowing through each resistor is different due its different resistance value. The total current I in the circuit divides to pass through the individual resistances.



According to Ohm’s Law
I = V/R

The current flowing through resistor R1 is I1   = V/R1
The current flowing through resistor R2 is I2 = V/R2
The current flowing through resistor R3 is I3     = V/R3

Total current     I = I1 + I2 + I3 = V/R1 + V/R2 + V/R3


V/R =  V(1/R1 + 1/R2 + 1/R3)
1/R =  1/R1 + 1/R2 + 1/R3

The reciprocal of equivalent resistance is equal to the sum of resistances which are connected in parallel.

When two resistor R1 and R2 are connected , then equivalent resistance is

R = R1R2/ (R1 + R2)

Resistors in Series - Parallel:

A series parallel circuit is shown in figure. Here the resistors R2 and R3 are connected in parallel. But the resistor R1 is connected in series with this parallel combination.

Equivalent resistance of the resistors R2 andR3 (parallel combination) is

1/ Rp = 1/R2 + 1/R3
Rp      = R2 R3 / (R2 + R3)
Total voltage

V = V1 + V2
    = IR1 + IRP
    = I ( R1 +RP )

IR = I ( R1 + RP )

R = R1 + RP
R = R1 + R2 R3/(R2  +  R3)

The current flowing through resistor R2 is     I2 = IR3/ (R2 + R3)
The current flowing through resistor R3 is     I3 = IR2/(R2 + R3)

Problem 1:

In the circuit shown in figure , Find the current , the voltage across each resistor and the power dissipated in each resistor.

Solution:

Total resistance, R = R1 + R2 + R3 = 6 + 9 + 10 = 25 Ω
I = V/R = 50/25 = 2 amps

Voltage drop across 6 Ω resistor, V1 = IR1 = 2 x 6 = 12 volts
Voltage drop across 9 Ω resistor, V2 = IR2 = 2 x 9 = 18 volts
Voltage drop across 10 Ω resistor, V3 = IR3 = 2 x 10 = 20 volts
Power dissipated in 6 Ω resistor, P1 = I2R1 = 4 x 6 = 24 watts
Power dissipated in 9 Ω resistor, P2 = I2R2 = 4 x 9 = 36 watts
Power dissipated in 10 Ω resistor, P3 = I2R3 = 4 x 10 = 40 watts

Problem 2:

Two resistors resistance 6 Ω and 12 Ω are connected in series. Find the equivalent Also find the equivalent resistance when they are connected in parallel.

Given data
Case 1: Series combination of 6 Ω and 12 Ω resistors.
R1 = 6 Ω, R2 = 12 Ω

Solution:

Equivalent resistance, R = R1 + R2 = 18 Ω
Case 2: Parallel combination of 6 Ω and 12 Ω resistors.
R1 = 6 Ω, R2 = 12 Ω



Equivalent resistance, 1/R = 1/ R1 + 1/R2
R = R1R2/(R1 + R2) = 6 x 12/(6 + 12) = 72/18 = 4 Ω

Problem 3:

In the circuit shown in figure , find the total resistance and current flowing through each branch.


Given data, R1 = 5 Ω, R2 = 10 Ω, R3 = 20 Ω

Solution:

Equivalent resistance, 1/R = 1/ R1 + 1/R2 + 1/R3 = 0.2 + 0.1 + 0.05 = 0.35
Total resistance, R = 2.86 Ω
The current flowing through 5 Ω resistor, I1 = V/R1 = 25/5 = 5 amps.
The current flowing through 10 Ω resistor, I2 = V/R2 = 25/10 = 2.5 amps
The current flowing through 20 Ω resistor, I3 = V/R3 = 25/20 = 1.25 amps

Problem 4:

Resistors of values 2, 3, 4, and 5 ohm are connected in parallel. If the total power absorbed by all the resistors is 200w, find the voltage applied to the circuit.
Power, P = 200 watts

R1 = 2 Ω, R2 = 3 Ω, R3 = 4 Ω, R4 = 5 Ω

Solution:

Total Resistance, 1/R = 1/ R1 + 1/R2 + 1/R3 + 1/R4 = 0.5 + 0.33 +0.25 + 0.2
1/R = 1.28,  R = 1/1.28 = 0.78 ohm
Power, P = V2/R
V2 = PX R
V = (P X R) = (200 X 0.78) = 12.5 volts

Problem 5:

Two resistors are connected in parallel across a 200V mains take a total current of 10amps. The power dissipated in one of the resistors is 1200. watts. Find the value of each resistor.

Given data:

Voltage, V = 200 volts
Current, I = 10 amps
Power dissipated in R1 of the resistors P1 = 1200 watts

Solution:

I1 = P1/V = 1200/ 200 = 6 amps.
Resistors, R1 = V/ I1 = 200/6 = 33.33 Ω
Total current, I = I1 + I2
10 = 6 + I2
I2 = 4 amps
Resistors, R2 = V/ I2 = 200/4 = 50 Ω

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