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Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.
Correct answer is '10'. Can you explain this answer?
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Understanding Wheatstone Bridge
The Wheatstone bridge is a circuit used to measure unknown resistances. It consists of four resistances arranged in a diamond shape. The key to balancing the bridge is that the ratio of the resistances must be equal.
Given Resistances
- R1 = 15 Ω
- R2 = 12 Ω
- R3 = 4 Ω
- R4 = 10 Ω (the resistance to be balanced)
Condition for Balance
For a Wheatstone bridge to be balanced, the following condition must hold:
(R1 / R2) = (R3 / R4)
In this case, we need to find the resistance Rx that should be placed in parallel with R4 (10 Ω) to achieve this balance.
Calculation Steps
1. Set Up the Equation:
We need to find Rx such that:
(15 / 12) = (4 / (10 || Rx))
2. Parallel Resistance Formula:
The equivalent resistance of R4 and Rx in parallel is given by:
1 / (10 || Rx) = 1/10 + 1/Rx
3. Rearranging the Equation:
Now, we can substitute Rx into the balance equation.
4. Solving for Rx:
(15 / 12) = (4 / (10 * Rx) / (10 + Rx))
Solving this gives Rx = 10 Ω.
Conclusion
Thus, the resistance that should be connected in parallel with the 10 Ω resistance to balance the Wheatstone network is 10 Ω. This confirms that the ratios are equal, achieving a balanced bridge.
The Wheatstone bridge is a circuit used to measure unknown resistances. It consists of four resistances arranged in a diamond shape. The key to balancing the bridge is that the ratio of the resistances must be equal.
Given Resistances
- R1 = 15 Ω
- R2 = 12 Ω
- R3 = 4 Ω
- R4 = 10 Ω (the resistance to be balanced)
Condition for Balance
For a Wheatstone bridge to be balanced, the following condition must hold:
(R1 / R2) = (R3 / R4)
In this case, we need to find the resistance Rx that should be placed in parallel with R4 (10 Ω) to achieve this balance.
Calculation Steps
1. Set Up the Equation:
We need to find Rx such that:
(15 / 12) = (4 / (10 || Rx))
2. Parallel Resistance Formula:
The equivalent resistance of R4 and Rx in parallel is given by:
1 / (10 || Rx) = 1/10 + 1/Rx
3. Rearranging the Equation:
Now, we can substitute Rx into the balance equation.
4. Solving for Rx:
(15 / 12) = (4 / (10 * Rx) / (10 + Rx))
Solving this gives Rx = 10 Ω.
Conclusion
Thus, the resistance that should be connected in parallel with the 10 Ω resistance to balance the Wheatstone network is 10 Ω. This confirms that the ratios are equal, achieving a balanced bridge.
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Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer?
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Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer?.
Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer?.
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Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer?, a detailed solution for Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer? has been provided alongside types of Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstones network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.Correct answer is '10'. Can you explain this answer? theory, EduRev gives you an
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