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A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this?
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A 1.5v cell of internal resistance 1ohm is connected to the resistors ...
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A 1.5v cell of internal resistance 1ohm is connected to the resistors ...
Analysis of the Circuit:
To solve the given problem, we need to analyze the circuit and use Ohm's law to calculate the current and potential difference across each resistor.
Given Data:
- Voltage of the cell (V) = 1.5 V
- Internal resistance of the cell (r) = 1 Ω
- Resistor 1 (R1) = 4 Ω
- Resistor 2 (R2) = 20 Ω
Calculations:
a) Current in the Circuit:
The total resistance in the circuit (R) can be calculated by adding the individual resistances in series.
R = R1 + R2
R = 4 Ω + 20 Ω
R = 24 Ω
We can now use Ohm's law to calculate the current (I) flowing in the circuit.
V = I * R
1.5 V = I * 24 Ω
Solving for I:
I = 1.5 V / 24 Ω
I ≈ 0.0625 A
Therefore, the current in the circuit is approximately 0.0625 A.
b) Potential Difference across Each Resistor:
To calculate the potential difference across each resistor, we can use Ohm's law again.
Potential difference across R1 (V1) = I * R1
V1 = 0.0625 A * 4 Ω
V1 = 0.25 V
Potential difference across R2 (V2) = I * R2
V2 = 0.0625 A * 20 Ω
V2 = 1.25 V
Therefore, the potential difference across R1 is 0.25 V and across R2 is 1.25 V.
c) Potential Difference across the Cell:
The potential difference across the cell (Vcell) can be calculated by subtracting the potential drop across the internal resistance from the cell voltage.
Vcell = V - (I * r)
Vcell = 1.5 V - (0.0625 A * 1 Ω)
Vcell = 1.5 V - 0.0625 V
Vcell ≈ 1.4375 V
Therefore, the potential difference across the cell is approximately 1.4375 V.
d) Voltage Drop when Current is Flowing:
The voltage drop when current is flowing can be calculated by multiplying the current with the resistance.
Voltage drop = I * R
Voltage drop = 0.0625 A * 24 Ω
Voltage drop = 1.5 V
Therefore, the voltage drop when current is flowing is 1.5 V.
Summary:
a) The current in the circuit is approximately 0.0625 A.
b) The potential difference across R1 is 0.25 V and across R2 is 1.25 V.
c) The potential difference across the cell is approximately 1.4375 V.
d) The voltage drop when current is flowing is 1.5 V.
To solve the given problem, we need to analyze the circuit and use Ohm's law to calculate the current and potential difference across each resistor.
Given Data:
- Voltage of the cell (V) = 1.5 V
- Internal resistance of the cell (r) = 1 Ω
- Resistor 1 (R1) = 4 Ω
- Resistor 2 (R2) = 20 Ω
Calculations:
a) Current in the Circuit:
The total resistance in the circuit (R) can be calculated by adding the individual resistances in series.
R = R1 + R2
R = 4 Ω + 20 Ω
R = 24 Ω
We can now use Ohm's law to calculate the current (I) flowing in the circuit.
V = I * R
1.5 V = I * 24 Ω
Solving for I:
I = 1.5 V / 24 Ω
I ≈ 0.0625 A
Therefore, the current in the circuit is approximately 0.0625 A.
b) Potential Difference across Each Resistor:
To calculate the potential difference across each resistor, we can use Ohm's law again.
Potential difference across R1 (V1) = I * R1
V1 = 0.0625 A * 4 Ω
V1 = 0.25 V
Potential difference across R2 (V2) = I * R2
V2 = 0.0625 A * 20 Ω
V2 = 1.25 V
Therefore, the potential difference across R1 is 0.25 V and across R2 is 1.25 V.
c) Potential Difference across the Cell:
The potential difference across the cell (Vcell) can be calculated by subtracting the potential drop across the internal resistance from the cell voltage.
Vcell = V - (I * r)
Vcell = 1.5 V - (0.0625 A * 1 Ω)
Vcell = 1.5 V - 0.0625 V
Vcell ≈ 1.4375 V
Therefore, the potential difference across the cell is approximately 1.4375 V.
d) Voltage Drop when Current is Flowing:
The voltage drop when current is flowing can be calculated by multiplying the current with the resistance.
Voltage drop = I * R
Voltage drop = 0.0625 A * 24 Ω
Voltage drop = 1.5 V
Therefore, the voltage drop when current is flowing is 1.5 V.
Summary:
a) The current in the circuit is approximately 0.0625 A.
b) The potential difference across R1 is 0.25 V and across R2 is 1.25 V.
c) The potential difference across the cell is approximately 1.4375 V.
d) The voltage drop when current is flowing is 1.5 V.
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A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this?
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A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this?.
A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this?.
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A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this?, a detailed solution for A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this? has been provided alongside types of A 1.5v cell of internal resistance 1ohm is connected to the resistors of 4ohm and 20ohm in series. Calculate a) current in the circuit. b)Potential difference across each resistor. c)Potential difference across cell and d) voltage drop when current is flowing. can anyone plz answer this? theory, EduRev gives you an
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