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Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire?
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Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and ...
Given information:
- Two cells with emf 1.5 V and 2 V
- Internal resistance of the cells: 2 ohm and 1 ohm
- Negative terminals of the cells are joined by a wire of 6 ohm
- Positive terminals of the cells are joined by a wire of 8 ohm
- The midpoints of these two wires are connected
Calculating the equivalent resistance:
To find the current and potential difference at the ends of the third wire, we need to calculate the equivalent resistance of the circuit.
1. Calculate the total resistance of the circuit:
- The internal resistance of the first cell is 2 ohm, and the wire resistance is 6 ohm, so the total resistance in that branch is 8 ohm.
- Similarly, the internal resistance of the second cell is 1 ohm, and the wire resistance is 8 ohm, so the total resistance in that branch is 9 ohm.
- Since the two branches are in parallel, we can calculate the total resistance using the formula: 1/Req = 1/R1 + 1/R2, where R1 = 8 ohm and R2 = 9 ohm.
- Solving the equation, we find Req = 4.5 ohm.
Calculating the current:
2. Calculate the total emf of the circuit:
- Since the emf of the first cell is 1.5 V and the emf of the second cell is 2 V, the total emf is the sum of these two: 1.5 V + 2 V = 3.5 V.
3. Calculate the current using Ohm's law:
- I = V/Req, where V is the total emf (3.5 V) and Req is the equivalent resistance (4.5 ohm).
- Solving the equation, we find I = 0.78 A (rounded to two decimal places).
Calculating the potential difference:
4. Calculate the potential difference at the ends of the third wire:
- Since the midpoints of the two wires are connected, the potential difference across the third wire is the same as the potential difference across the second wire.
- To calculate the potential difference across the second wire, we can use Ohm's law: V = I * R, where I is the current (0.78 A) and R is the resistance of the second wire (8 ohm).
- Solving the equation, we find V = 6.24 V (rounded to two decimal places).
Summary:
- The current in the circuit is 0.78 A.
- The potential difference at the ends of the third wire is 6.24 V.
- Two cells with emf 1.5 V and 2 V
- Internal resistance of the cells: 2 ohm and 1 ohm
- Negative terminals of the cells are joined by a wire of 6 ohm
- Positive terminals of the cells are joined by a wire of 8 ohm
- The midpoints of these two wires are connected
Calculating the equivalent resistance:
To find the current and potential difference at the ends of the third wire, we need to calculate the equivalent resistance of the circuit.
1. Calculate the total resistance of the circuit:
- The internal resistance of the first cell is 2 ohm, and the wire resistance is 6 ohm, so the total resistance in that branch is 8 ohm.
- Similarly, the internal resistance of the second cell is 1 ohm, and the wire resistance is 8 ohm, so the total resistance in that branch is 9 ohm.
- Since the two branches are in parallel, we can calculate the total resistance using the formula: 1/Req = 1/R1 + 1/R2, where R1 = 8 ohm and R2 = 9 ohm.
- Solving the equation, we find Req = 4.5 ohm.
Calculating the current:
2. Calculate the total emf of the circuit:
- Since the emf of the first cell is 1.5 V and the emf of the second cell is 2 V, the total emf is the sum of these two: 1.5 V + 2 V = 3.5 V.
3. Calculate the current using Ohm's law:
- I = V/Req, where V is the total emf (3.5 V) and Req is the equivalent resistance (4.5 ohm).
- Solving the equation, we find I = 0.78 A (rounded to two decimal places).
Calculating the potential difference:
4. Calculate the potential difference at the ends of the third wire:
- Since the midpoints of the two wires are connected, the potential difference across the third wire is the same as the potential difference across the second wire.
- To calculate the potential difference across the second wire, we can use Ohm's law: V = I * R, where I is the current (0.78 A) and R is the resistance of the second wire (8 ohm).
- Solving the equation, we find V = 6.24 V (rounded to two decimal places).
Summary:
- The current in the circuit is 0.78 A.
- The potential difference at the ends of the third wire is 6.24 V.
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Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire?
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Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire?.
Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire?.
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Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire?, a detailed solution for Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire? has been provided alongside types of Two cells of emf 1.5 volt and 2 volt and internal resitance 2 ohm and 1 ohm respectively, have their negative terminals joined by a wire of 6ohm and positive terminal by another wire of 8 ohm connect the mid points of these two wire. Fins the current and potential difference at the ends of the third wire? theory, EduRev gives you an
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