To check who made the correct calculation, let's compare both answers using the current and time. 1 A's calculation: The current I. = 5A, and the charge flowing in 1 minute (60 seconds) is given by: Q = I × t = 5A × 60s = 300 C So, A's calculation is correct. 2. B's calculation: The number of electrRead more
To check who made the correct calculation, let’s compare both answers using the current and time. 1
A’s calculation:
The current
I. = 5A, and the charge flowing in 1 minute (60 seconds) is given by:
Q = I × t = 5A × 60s = 300 C
So, A’s calculation is correct.
2. B’s calculation:
The number of electrons flowing per second can be calculated by:
Number of electrons = Q/e
where e = 1.6 ×10⁻¹⁹C (charge of one electron) and
Q = 5C/s.
Number of electrons = 5/1.6 × 10⁻¹⁹ 3.125 × 10¹⁹ electrons/s
So, B’s calculation is also correct.
Thus, the correct answer is (c) both A and B.
In a potentiometer experiment, the internal resistance of the cell can be determined using the formula: L₁/L₂ = r + R/r where L₁ is the original balancing length, L₂ is the new balancing length, r is the internal resistance, and R is the external resistance. Substituting the values, we get r = 3Ω. TRead more
In a potentiometer experiment, the internal resistance of the cell can be determined using the formula:
L₁/L₂ = r + R/r
where
L₁ is the original balancing length, L₂ is the new balancing length,
r is the internal resistance, and R is the external resistance. Substituting the values, we get
r = 3Ω. Therefore, the internal resistance is 3 Ω. The correct answer is (c) 3 Ω.
The total resistance in the circuit is Rcell+Rwire = 3Ω + 3Ω = 6Ω The current is I = 10V/6Ω = 5/3 A. The potential gradient is V/L = IR/500 = 5/3 3/500 = 10 mV/cm. Answer: (b). For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
The total resistance in the circuit is Rcell+Rwire = 3Ω + 3Ω = 6Ω
The current is I = 10V/6Ω = 5/3 A. The potential gradient is V/L = IR/500 = 5/3 3/500 = 10 mV/cm. Answer: (b).
Let P₁ and P₂ be powers across R₁ and R₂. In series, power is divided inversely as resistance. In parallel, power is directly proportional to conductance. Using given conditions and solving equations, the ratio P₁ /P₂ = 4. Answer: (b) 4. For more visit here: https://www.tiwariacademy.com/ncert-solutRead more
Let P₁ and P₂ be powers across R₁ and R₂. In series, power is divided inversely as resistance. In parallel, power is directly proportional to conductance. Using given conditions and solving equations, the ratio P₁ /P₂ = 4. Answer: (b) 4.
Two students A and B calculate the charge flowing through a circuit. A concludes that 300 C of charge flows in 1 minute. B concludes that 3.125 x 10¹⁹ electrons flow in 1 second. If the current measured in the circuit is 5 A, then the correct calculation is done by
To check who made the correct calculation, let's compare both answers using the current and time. 1 A's calculation: The current I. = 5A, and the charge flowing in 1 minute (60 seconds) is given by: Q = I × t = 5A × 60s = 300 C So, A's calculation is correct. 2. B's calculation: The number of electrRead more
To check who made the correct calculation, let’s compare both answers using the current and time. 1
A’s calculation:
The current
I. = 5A, and the charge flowing in 1 minute (60 seconds) is given by:
Q = I × t = 5A × 60s = 300 C
So, A’s calculation is correct.
2. B’s calculation:
The number of electrons flowing per second can be calculated by:
Number of electrons = Q/e
where e = 1.6 ×10⁻¹⁹C (charge of one electron) and
Q = 5C/s.
Number of electrons = 5/1.6 × 10⁻¹⁹ 3.125 × 10¹⁹ electrons/s
So, B’s calculation is also correct.
Thus, the correct answer is (c) both A and B.
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A battery of 15 V and negligible internal resistance is connected across a 50 Ω resistor. The amount of energy dissipated as heat in the resistor in one minute is :
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In a potentiometer experiment, the balancing length with a cell is 120 cm. When the cell is shunted by a 1Ω resistance, the balancing length becomes 40 cm. The internal resistance of the cell is :
In a potentiometer experiment, the internal resistance of the cell can be determined using the formula: L₁/L₂ = r + R/r where L₁ is the original balancing length, L₂ is the new balancing length, r is the internal resistance, and R is the external resistance. Substituting the values, we get r = 3Ω. TRead more
In a potentiometer experiment, the internal resistance of the cell can be determined using the formula:
L₁/L₂ = r + R/r
where
L₁ is the original balancing length, L₂ is the new balancing length,
r is the internal resistance, and R is the external resistance. Substituting the values, we get
r = 3Ω. Therefore, the internal resistance is 3 Ω. The correct answer is (c) 3 Ω.
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A cell of internal resistance 3 Ω and emf 10 V is connected to a uniform wire of length 500 cm and resistance 3 Ω. The potential gradient in the wire is
The total resistance in the circuit is Rcell+Rwire = 3Ω + 3Ω = 6Ω The current is I = 10V/6Ω = 5/3 A. The potential gradient is V/L = IR/500 = 5/3 3/500 = 10 mV/cm. Answer: (b). For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
The total resistance in the circuit is Rcell+Rwire = 3Ω + 3Ω = 6Ω
The current is I = 10V/6Ω = 5/3 A. The potential gradient is V/L = IR/500 = 5/3 3/500 = 10 mV/cm. Answer: (b).
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When two resistances R₁ and R₂ are connected in series, they consume 12 W power. When they are connected in parallel, they consume 50 W power. What is the ratio of the Power of R₁ and R₂?
Let P₁ and P₂ be powers across R₁ and R₂. In series, power is divided inversely as resistance. In parallel, power is directly proportional to conductance. Using given conditions and solving equations, the ratio P₁ /P₂ = 4. Answer: (b) 4. For more visit here: https://www.tiwariacademy.com/ncert-solutRead more
Let P₁ and P₂ be powers across R₁ and R₂. In series, power is divided inversely as resistance. In parallel, power is directly proportional to conductance. Using given conditions and solving equations, the ratio P₁ /P₂ = 4. Answer: (b) 4.
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