Question

In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one dominates. Make some speci c examples of resistorsin parallel and in series to explain this rule of thumb.

Answer 1

Answer:

 In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one dominates. Make some specific examples of resistorsin parallel and in series to explain this rule of thumb.

Explanation:

Lets start with Resistors in series:

Suppose a 12v battery is connected in a circuit with 3 resistors in series.

Voltage = 12 V

R1          = 1 ohm

R2         = 6 ohm

R3         = 13 ohm

Total resistance is simply sum of all resistors connected in series as following:

Rs   = R1 + R2 + R3

Rs   = 1 + 6 + 13

Rs   = 20 ohm

Here Rs = Total Resistance in series circuit

Now total current can be found by using ohm's law V = IR

I = V/R

Here R = R total

I = 12/20 = 0.6 Ampere

In series circuit current remains same.

so I1 = I2 = I3 = I

Now we can find Power dissipation across every resistor to show the dominant Resistor as:

P1 = I² R1 = (0.6 A)² ( 1 ohm) = 0.36 watt.          (a)

P2 = I² R2 = (0.6 A)² ( 6 ohm) = 2.16 watt.        (b)

P3 = I² R3 = (0.6 A)² ( 13 ohm) = 4.68 watt.      (c)

From equation a, b and c  we can conclude that more power is dissipated across R3 which is the Larger than both other resistors R1 and R2.

Now lets take the case of Parallel Circuit with same Values of Voltage and resistors But now in parallel.

In parallel circuit Total resistance is calculated as following:

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/1 + 1/6 + 1/13 = 1 + 0.1667 + 0.0769

1/Rp = 1.2436

Rp   = 0.8041 ohm

Now total current is as following:

I = V/R

I = 12/0.8041 = 14.92 A

Now we calculate individual currents :

In parallel circuit Voltage remains same.

I1 = V/R1

I1 = 12/1 = 12 A

Similarly,

I2 = V/R2

I2 = 12/ 6 = 2 A

and

I3 = V/R3

I3 = 12/13 = 0.92 A

Now Power dissipation can be calculated as :

P1 = I1² * R1

P1 = 12² * 1 = 144 * 1

P1 = 144 Watt            (1)

Similarly,

P2 = I2² * R2

P2 = 2² * 6 = 4 * 6

P2 = 24 Watt            (2)

And

P3 = I3² * R3

P3 = (0.92)² * 13 = 0.8464 * 13

P3 = 11 Watt.              (3)

Now from Equation 1, 2 and 3 we can conclude that More power is dissipated across Lower resistor .

Hence in a circuit with parallel resistors, the smaller resistance dominates; in a circuit with resistors in series, the larger one dominates.