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Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.
Correct answer is '30'. Can you explain this answer?
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Two batteries (connected in series) of same emf 10 V of internal resis...
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And 30 ohms respectively are connected to a 50 ohm resistor. Find the current flowing through the circuit and the voltage drop across each battery.
To solve the problem, we first need to calculate the total internal resistance of the battery circuit. The internal resistances of the two batteries are connected in series, so they add up to give:
R_total = R1 + R2 = 20 + 30 = 50 ohms
The total resistance of the circuit is the sum of the internal resistance and the external resistor:
R = R_total + R_ext = 50 + 50 = 100 ohms
Using Ohm's law, we can calculate the current flowing through the circuit:
I = V / R = 10 / 100 = 0.1 A
The voltage drop across the external resistor can be calculated as:
V_ext = I * R_ext = 0.1 * 50 = 5 V
Since the batteries are connected in series, the total voltage drop across the circuit is the sum of the voltages of the two batteries:
V_total = V1 + V2 = 10 + 10 = 20 V
We can use the voltage divider rule to calculate the voltage drop across each battery:
V1 = V_total * (R1 / R_total) = 20 * (20 / 50) = 8 V
V2 = V_total * (R2 / R_total) = 20 * (30 / 50) = 12 V
Therefore, the current flowing through the circuit is 0.1 A, and the voltage drop across the two batteries are 8 V and 12 V respectively.
To solve the problem, we first need to calculate the total internal resistance of the battery circuit. The internal resistances of the two batteries are connected in series, so they add up to give:
R_total = R1 + R2 = 20 + 30 = 50 ohms
The total resistance of the circuit is the sum of the internal resistance and the external resistor:
R = R_total + R_ext = 50 + 50 = 100 ohms
Using Ohm's law, we can calculate the current flowing through the circuit:
I = V / R = 10 / 100 = 0.1 A
The voltage drop across the external resistor can be calculated as:
V_ext = I * R_ext = 0.1 * 50 = 5 V
Since the batteries are connected in series, the total voltage drop across the circuit is the sum of the voltages of the two batteries:
V_total = V1 + V2 = 10 + 10 = 20 V
We can use the voltage divider rule to calculate the voltage drop across each battery:
V1 = V_total * (R1 / R_total) = 20 * (20 / 50) = 8 V
V2 = V_total * (R2 / R_total) = 20 * (30 / 50) = 12 V
Therefore, the current flowing through the circuit is 0.1 A, and the voltage drop across the two batteries are 8 V and 12 V respectively.
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Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. Can you explain this answer?
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Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. 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 Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. Can you explain this answer?.
Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. 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 Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two batteries (connected in series) of same emf 10 V of internal resistances 20Ω and 5Ω are connected to a load resistance of 30Ω. Now an unknown resistance x is connected in parallel to the load resistance. Find value of x so that potential drop of battery having internal resistance 20Ω becomes zero.Correct answer is '30'. Can you explain this answer?.
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