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When two identical resistors are connected in series across a battery, the power dissipated is 10 W. If these resistors are connected in parallel across the same battery, the total power dissipated will be
- a)10 W
- b)20 W
- c)40 W
- d)80 W
Correct answer is option 'C'. Can you explain this answer?
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When two identical resistors are connected in series across a battery,...
Solution:
When two identical resistors are connected in series across a battery, the power dissipated is 10 W.
Let the resistance of each resistor be R.
The total resistance in series connection is given by:
R_series = R + R = 2R
The power dissipated is given by:
P = V^2/R_series
where V is the voltage of the battery.
Substituting the given values, we get:
10 = V^2/(2R)
On solving, we get:
V = √(20R)
Now, these resistors are connected in parallel across the same battery.
The total resistance in parallel connection is given by:
1/R_parallel = 1/R + 1/R = 2/R
R_parallel = R/2
The power dissipated is given by:
P = V^2/R_parallel
Substituting the value of V and R_parallel, we get:
P = 20R/(R/2) = 40 W
Hence, the total power dissipated when the two identical resistors are connected in parallel across the same battery is 40 W.
Therefore, option C is the correct answer.
When two identical resistors are connected in series across a battery, the power dissipated is 10 W.
Let the resistance of each resistor be R.
The total resistance in series connection is given by:
R_series = R + R = 2R
The power dissipated is given by:
P = V^2/R_series
where V is the voltage of the battery.
Substituting the given values, we get:
10 = V^2/(2R)
On solving, we get:
V = √(20R)
Now, these resistors are connected in parallel across the same battery.
The total resistance in parallel connection is given by:
1/R_parallel = 1/R + 1/R = 2/R
R_parallel = R/2
The power dissipated is given by:
P = V^2/R_parallel
Substituting the value of V and R_parallel, we get:
P = 20R/(R/2) = 40 W
Hence, the total power dissipated when the two identical resistors are connected in parallel across the same battery is 40 W.
Therefore, option C is the correct answer.
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Community Answer
When two identical resistors are connected in series across a battery,...
When two or more electrical devices of the same voltage are connected in parallel connection, their equivalent power rating can be calculated as
Peq = P1 + P2 + P3 + ……
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in parallel connection, their equivalent power rating can be calculated as
Peq = nP
When two or more electrical devices of the same voltage rating are connected in a series connection, their equivalent power rating can be calculated as
Peq = P1 + P2 + P3 + ……
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in parallel connection, their equivalent power rating can be calculated as
Peq = nP
When two or more electrical devices of the same voltage rating are connected in a series connection, their equivalent power rating can be calculated as
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in series connection, their equivalent power rating can be calculated as
Calculation:
When two identical bulbs are resistors are connected in series across a battery, their equivalent power (Peq) is ‘10 W’
∴ The power rating of each bulb, P = 20 W
When two identical bulbs are resistors are connected in series across a battery, their equivalent power (Peq) is ‘10 W’
∴ The power rating of each bulb, P = 20 W
When these two resistors are connected in parallel, then the equivalent power dissipated is
Peq = 2 × P
Peq = 40 W
Peq = 2 × P
Peq = 40 W
Alternate Approach:
Series circuit:
In the series circuit, the voltage drop across each resistor = V/2
The total current through the circuit = V/2R amps.
The power dissipated by each resistor = V/2 × V/2R = V2/4R
Which must be equal to 5 Watts in order to make a total of 10 W.
Series circuit:
In the series circuit, the voltage drop across each resistor = V/2
The total current through the circuit = V/2R amps.
The power dissipated by each resistor = V/2 × V/2R = V2/4R
Which must be equal to 5 Watts in order to make a total of 10 W.
Parallel circuit:
The voltage drop across each resistor = V.
The current through each resistor = V/R.
Multiplying these gives V2/R which is 4 times the series value of V2/4R
∴ The total power dissipated in parallel circuit = 4 × 10 = 40 W
The voltage drop across each resistor = V.
The current through each resistor = V/R.
Multiplying these gives V2/R which is 4 times the series value of V2/4R
∴ The total power dissipated in parallel circuit = 4 × 10 = 40 W
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