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Range of Power for Parallel Resistors

When replacing a single resistor with an equivalent pair of resistors in parallel, the total resistance of the combination will be the same as the original resistor.However, power handling capability of the new parallel combination can differ from that of the original resistor. Consider this:

Total Resistance Calculation: For two resistors, R1 R_1 and R2 R_2 , in parallel, the equivalent resistance Req R_{eq} is given by:

1Req=1R1+1R2 \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
parallel resistor picture

Power Distribution: The power dissipated in a resistor is given by P=V2R P = \frac{V^2}{R} , where V V is the voltage across the resistor. For the parallel combination, each resistor has same voltage across them.

Power Sharing: If P1 P_1 and P2 P_2 are the powers dissipated by resistors R1 R_1 and R2 R_2 respectively, then:

P1=V2R1andP2=V2R2 P_1 = \frac{V^2}{R_1} \quad \text{and} \quad P_2 = \frac{V^2}{R_2}

The total power dissipated by the parallel combination Ptotal P_{total} is:

Ptotal=P1+P2 P_{total} = P_1 + P_2

Power Rating of Individual Resistors: The power rating of the new resistors in parallel must be selected such that neither of them exceeds its maximum rated power. Let's suppose that the original resistor had a power rating Poriginal P_{original} .

Minimum Power Rating: The minimum power rating of each resistor in the parallel combination must be sufficient to handle its share of the total power. For equal resistors (R1=R2 R_1 = R_2 ):

P1=P2=Ptotal2 P_1 = P_2 = \frac{P_{total}}{2}

Therefore, each resistor must have a power rating of at least Poriginal2 \frac{P_{original}}{2} .

General Case: For resistors of different values, the power rating of each resistor has to be determined considering the ratio of their resistances. For resistors R1 R_1 and R2 R_2 :

P1=R2R1+R2Poriginaland P_1 = \frac{R_2}{R_1 + R_2} P_{original} \quad \text{and} P2=R1R1+R2Poriginal P_2 = \frac{R_1}{R_1 + R_2} P_{original}

Example: 150 Ω and 180 Ω Resistors in Parallel

Let's analyze an example where a 150 Ω resistor is in parallel with another of 180 Ω. We will calculate the equivalent resistance and the power ratings for each resistor.

Total Resistance Calculation: For two resistors, R1=150Ω R_1 = 150 \, \Omega and R2=180Ω R_2 = 180 \, \Omega , in parallel, the equivalent resistance Req R_{eq} is given by:

1Req=1R1+1R2=1150+1180 \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{150} + \frac{1}{180} 1Req=180+150150×180=33027000 \frac{1}{R_{eq}} = \frac{180 + 150}{150 \times 180} = \frac{330}{27000} Req=2700033081.82Ω R_{eq} = \frac{27000}{330} \approx 81.82 \, \Omega

Power Distribution: If the total power applied is Ptotal P_{total} , the power dissipated in each resistor can be found using:

P1=R2R1+R2PtotalandP2=R1R1+R2Ptotal P_1 = \frac{R_2}{R_1 + R_2} P_{total} \quad \text{and} \quad P_2 = \frac{R_1}{R_1 + R_2} P_{total}

Example Calculation: Consiter that the total power Ptotal P_{total} is 10 W.

P1=180150+180×10W=180330×10W5.45W P_1 = \frac{180}{150 + 180} \times 10 \, \text{W} = \frac{180}{330} \times 10 \, \text{W} \approx 5.45 \, \text{W} P2=150150+180×10W=150330×10W4.55W P_2 = \frac{150}{150 + 180} \times 10 \, \text{W} = \frac{150}{330} \times 10 \, \text{W} \approx 4.55 \, \text{W}

Power Rating of Individual Resistors: Each resistor must have a power rating that can handle at least its share of the total power. In this example:

P15.45W(for the 150 Ω resistor) P_1 \approx 5.45 \, \text{W} \quad \text{(for the 150 Ω resistor)} P24.55W(for the 180 Ω resistor) P_2 \approx 4.55 \, \text{W} \quad \text{(for the 180 Ω resistor)}
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