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Three equal resistors are connected in series, across a source of emf, dissipate 10 W of power. What would be the power dissipated when they are connected in parallel across the same source.
- a)10 W
- b)30 W
- c)90 W
- d)270 W
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Three equal resistors are connected in series, across a source of emf,...
Let the resistance of each resistor = R Ω
If three equal resistors are connected in series (R1) = 3R
Power dissipated in this case (P1) = 10 W
If three equal resistors are connected in parallel (R2) = R/3
Power dissipated in this case = P2
We know that,
For the same voltage, power is inversely proportional to resistance.
From the first case to the second case, the resistance decreases by 9 times. Hence the power will increase by 9 times.
Power in the second case (P2) = 90 W
If three equal resistors are connected in series (R1) = 3R
Power dissipated in this case (P1) = 10 W
If three equal resistors are connected in parallel (R2) = R/3
Power dissipated in this case = P2
We know that,
For the same voltage, power is inversely proportional to resistance.
From the first case to the second case, the resistance decreases by 9 times. Hence the power will increase by 9 times.
Power in the second case (P2) = 90 W
Most Upvoted Answer
Three equal resistors are connected in series, across a source of emf,...
Given:
- Three equal resistors are connected in series.
- The resistors are connected across a source of emf.
- The power dissipated is 10 W.
To find:
The power dissipated when the resistors are connected in parallel across the same source.
Solution:
1. Power dissipated in series connection:
When resistors are connected in series, the total resistance is the sum of individual resistances.
Formula:
For resistors connected in series, the total resistance (R) is given by:
R = R1 + R2 + R3 + ...
The power dissipated (P) is given by:
P = (V^2) / R
Where:
V is the voltage across the resistors.
Explanation:
In a series connection, the voltage across each resistor is the same.
Let's assume the resistance of each resistor is RΩ.
Since the resistors are equal, the total resistance (R) is given by:
R = R1 + R2 + R3 = 3R
The power dissipated (P1) in the series connection is given by:
P1 = (V^2) / R = (V^2) / (3R)
Given that P1 = 10 W, we can write the equation as:
10 = (V^2) / (3R)
Simplifying the equation, we get:
V^2 = 30R
2. Power dissipated in parallel connection:
When resistors are connected in parallel, the total resistance is given by the reciprocal of the sum of the reciprocals of individual resistances.
Formula:
For resistors connected in parallel, the total resistance (R) is given by:
1/R = 1/R1 + 1/R2 + 1/R3 + ...
The power dissipated (P2) is given by:
P2 = (V^2) / R
Where:
V is the voltage across the resistors.
Explanation:
In a parallel connection, the voltage across each resistor is the same.
Using the assumption that the resistance of each resistor is RΩ, the total resistance (R) is given by:
1/R = 1/R1 + 1/R2 + 1/R3 = 1/R + 1/R + 1/R = 3/R
Simplifying the equation, we get:
1/R = 3/R
Therefore, the total resistance (R) in the parallel connection is R/3.
The power dissipated (P2) in the parallel connection is given by:
P2 = (V^2) / (R/3) = 3(V^2) / R
Since P2 = 3P1, we can conclude that the power dissipated in the parallel connection is three times the power dissipated in the series connection.
Therefore, the power dissipated when the resistors are connected in parallel across the same source is 3 * 10 W = 30 W.
Hence, the correct answer is option 'b) 30 W'.
- Three equal resistors are connected in series.
- The resistors are connected across a source of emf.
- The power dissipated is 10 W.
To find:
The power dissipated when the resistors are connected in parallel across the same source.
Solution:
1. Power dissipated in series connection:
When resistors are connected in series, the total resistance is the sum of individual resistances.
Formula:
For resistors connected in series, the total resistance (R) is given by:
R = R1 + R2 + R3 + ...
The power dissipated (P) is given by:
P = (V^2) / R
Where:
V is the voltage across the resistors.
Explanation:
In a series connection, the voltage across each resistor is the same.
Let's assume the resistance of each resistor is RΩ.
Since the resistors are equal, the total resistance (R) is given by:
R = R1 + R2 + R3 = 3R
The power dissipated (P1) in the series connection is given by:
P1 = (V^2) / R = (V^2) / (3R)
Given that P1 = 10 W, we can write the equation as:
10 = (V^2) / (3R)
Simplifying the equation, we get:
V^2 = 30R
2. Power dissipated in parallel connection:
When resistors are connected in parallel, the total resistance is given by the reciprocal of the sum of the reciprocals of individual resistances.
Formula:
For resistors connected in parallel, the total resistance (R) is given by:
1/R = 1/R1 + 1/R2 + 1/R3 + ...
The power dissipated (P2) is given by:
P2 = (V^2) / R
Where:
V is the voltage across the resistors.
Explanation:
In a parallel connection, the voltage across each resistor is the same.
Using the assumption that the resistance of each resistor is RΩ, the total resistance (R) is given by:
1/R = 1/R1 + 1/R2 + 1/R3 = 1/R + 1/R + 1/R = 3/R
Simplifying the equation, we get:
1/R = 3/R
Therefore, the total resistance (R) in the parallel connection is R/3.
The power dissipated (P2) in the parallel connection is given by:
P2 = (V^2) / (R/3) = 3(V^2) / R
Since P2 = 3P1, we can conclude that the power dissipated in the parallel connection is three times the power dissipated in the series connection.
Therefore, the power dissipated when the resistors are connected in parallel across the same source is 3 * 10 W = 30 W.
Hence, the correct answer is option 'b) 30 W'.
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Three equal resistors are connected in series, across a source of emf, dissipate 10 W of power. What would be the power dissipated when they are connected in parallel across the same source.a)10 Wb)30 Wc)90 Wd)270 WCorrect answer is option 'C'. Can you explain this answer?
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