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In a network made up of linear resistors and ideal voltage sources, values of all resistors are doubled. Then the voltage across each resistor is
- a)doubled
- b)halved
- c)Decreases four times
- d)Constant
Correct answer is option 'D'. Can you explain this answer?
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In a network made up of linear resistors and ideal voltage sources, va...
Even on changing the values of linear resistors, the voltage remains constant in case of ideal voltage source.
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In a network made up of linear resistors and ideal voltage sources, va...
I think that 'D' should be the correct answer and power loss will be 1/4
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In a network made up of linear resistors and ideal voltage sources, va...
Explanation:
When the values of all resistors in a network are doubled, the voltage across each resistor remains constant.
Reasoning:
To understand why this happens, let's consider a simple circuit with a single resistor connected to an ideal voltage source.
Case 1: Original Circuit
In the original circuit, let's assume that the resistor has a resistance of R and the voltage source provides a voltage of V. According to Ohm's law, the current flowing through the resistor is given by I = V/R.
The voltage across the resistor can be determined using Ohm's law as well. The voltage drop across the resistor (VR) is given by VR = IR = (V/R) * R = V.
Therefore, in the original circuit, the voltage across the resistor is V.
Case 2: Doubled Resistors
Now, let's consider the case where all the resistors in the network are doubled. In this case, the resistance of the resistor in our circuit becomes 2R.
The voltage source remains the same, providing a voltage of V.
Using Ohm's law, the current flowing through the resistor is now given by I = V/(2R).
The voltage drop across the resistor is VR = IR = (V/(2R)) * (2R) = V.
Therefore, even though the resistance of the resistor doubled, the voltage across it remains the same.
Conclusion:
When the values of all resistors in a network are doubled, the voltage across each resistor remains constant. This can be observed from the fact that the voltage drop across a resistor only depends on the current flowing through it and not on the resistance itself.
When the values of all resistors in a network are doubled, the voltage across each resistor remains constant.
Reasoning:
To understand why this happens, let's consider a simple circuit with a single resistor connected to an ideal voltage source.
Case 1: Original Circuit
In the original circuit, let's assume that the resistor has a resistance of R and the voltage source provides a voltage of V. According to Ohm's law, the current flowing through the resistor is given by I = V/R.
The voltage across the resistor can be determined using Ohm's law as well. The voltage drop across the resistor (VR) is given by VR = IR = (V/R) * R = V.
Therefore, in the original circuit, the voltage across the resistor is V.
Case 2: Doubled Resistors
Now, let's consider the case where all the resistors in the network are doubled. In this case, the resistance of the resistor in our circuit becomes 2R.
The voltage source remains the same, providing a voltage of V.
Using Ohm's law, the current flowing through the resistor is now given by I = V/(2R).
The voltage drop across the resistor is VR = IR = (V/(2R)) * (2R) = V.
Therefore, even though the resistance of the resistor doubled, the voltage across it remains the same.
Conclusion:
When the values of all resistors in a network are doubled, the voltage across each resistor remains constant. This can be observed from the fact that the voltage drop across a resistor only depends on the current flowing through it and not on the resistance itself.
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In a network made up of linear resistors and ideal voltage sources, values of all resistors are doubled. Then the voltage across each resistor isa)doubledb)halvedc)Decreases four timesd)ConstantCorrect answer is option 'D'. Can you explain this answer?
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