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When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) is
Correct answer is '4'. Can you explain this answer?
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When two identical batteries of internal resistance 1 Ω each are conn...
In series,
Rate of heat produced in R is
In parallel,
Rate of heat produced in R is
According to the given problem, J1 = 2.25J2
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When two identical batteries of internal resistance 1 Ω each are conn...
To solve this problem, let's analyze the two scenarios separately: when the batteries are connected in series and when they are connected in parallel.
1. Batteries connected in series:
- When the batteries are connected in series, their voltages add up while their internal resistances also add up.
- Let's assume that the voltage across each battery is V (since they are identical), and the total voltage across the combination is V_total = 2V.
- The total internal resistance of the combination is R_total = 2Ω.
- The equivalent resistance of the circuit (including the external resistor R) is R_eq = R + R_total = R + 2Ω.
- Using Ohm's Law (V = IR), the current passing through the circuit is I = V_total / R_eq = 2V / (R + 2Ω).
- The rate of heat produced in the resistor is given by Joule's law as J1 = I^2 * R = (2V / (R + 2Ω))^2 * R = 4V^2R / (R + 2Ω)^2.
2. Batteries connected in parallel:
- When the batteries are connected in parallel, their voltages remain the same while their internal resistances decrease.
- The total internal resistance of the combination is R_total = 1Ω / 2 = 0.5Ω.
- The equivalent resistance of the circuit (including the external resistor R) is R_eq = 1 / (1/R + 1/R_total) = 1 / (1/R + 1/0.5) = 1 / (2/R + 1).
- The current passing through the circuit is I = V / R_eq = V / (2/R + 1).
- The rate of heat produced in the resistor is J2 = I^2 * R = (V / (2/R + 1))^2 * R = V^2R / (4/R^2 + 4/R + 1).
Given that J1 = 2.25J and J2 = J1 / 2 = 1.125J, we can set up the following equation:
4V^2R / (R + 2Ω)^2 = V^2R / (4/R^2 + 4/R + 1)
Simplifying this equation, we get:
4(R^2 + 4R + 4) = R(4R + 8)
4R^2 + 16R + 16 = 4R^2 + 8R
8R = 16
R = 2Ω
However, this is not the correct answer. We made a mistake assuming that J2 = J1 / 2. We need to consider that the power dissipated in the resistor is proportional to the square of the current passing through it. Therefore, J2 = J1 / 4 = 0.5625J.
Now, let's solve the equation again:
4V^2R / (R + 2Ω)^2 = V^2R / (4/R^2 + 4/R + 1)
4(R^2 + 4R + 4) = R(4R + 8)
4R^2 + 16R + 16 =
1. Batteries connected in series:
- When the batteries are connected in series, their voltages add up while their internal resistances also add up.
- Let's assume that the voltage across each battery is V (since they are identical), and the total voltage across the combination is V_total = 2V.
- The total internal resistance of the combination is R_total = 2Ω.
- The equivalent resistance of the circuit (including the external resistor R) is R_eq = R + R_total = R + 2Ω.
- Using Ohm's Law (V = IR), the current passing through the circuit is I = V_total / R_eq = 2V / (R + 2Ω).
- The rate of heat produced in the resistor is given by Joule's law as J1 = I^2 * R = (2V / (R + 2Ω))^2 * R = 4V^2R / (R + 2Ω)^2.
2. Batteries connected in parallel:
- When the batteries are connected in parallel, their voltages remain the same while their internal resistances decrease.
- The total internal resistance of the combination is R_total = 1Ω / 2 = 0.5Ω.
- The equivalent resistance of the circuit (including the external resistor R) is R_eq = 1 / (1/R + 1/R_total) = 1 / (1/R + 1/0.5) = 1 / (2/R + 1).
- The current passing through the circuit is I = V / R_eq = V / (2/R + 1).
- The rate of heat produced in the resistor is J2 = I^2 * R = (V / (2/R + 1))^2 * R = V^2R / (4/R^2 + 4/R + 1).
Given that J1 = 2.25J and J2 = J1 / 2 = 1.125J, we can set up the following equation:
4V^2R / (R + 2Ω)^2 = V^2R / (4/R^2 + 4/R + 1)
Simplifying this equation, we get:
4(R^2 + 4R + 4) = R(4R + 8)
4R^2 + 16R + 16 = 4R^2 + 8R
8R = 16
R = 2Ω
However, this is not the correct answer. We made a mistake assuming that J2 = J1 / 2. We need to consider that the power dissipated in the resistor is proportional to the square of the current passing through it. Therefore, J2 = J1 / 4 = 0.5625J.
Now, let's solve the equation again:
4V^2R / (R + 2Ω)^2 = V^2R / (4/R^2 + 4/R + 1)
4(R^2 + 4R + 4) = R(4R + 8)
4R^2 + 16R + 16 =
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When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer?
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When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. 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 When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer?.
When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. 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 When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer?.
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When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer?, a detailed solution for When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer? has been provided alongside types of When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J,, then the value of R (in Ω) isCorrect answer is '4'. Can you explain this answer? theory, EduRev gives you an
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