The Question is from neet-physics current-electricity , and it was created by the author SUBHANI

Question

The charge flowing through a resistance R varies with time t as ${\rm{Q}}\,{\rm{ = }}\,{\rm{at}}\,{\rm{ - b}}{{\rm{t}}^2}$  where a and b are positive constants. The total heat produced in R is:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Heat generated in a resistor due to current flow
	**Related Topics**: Calculus (differentiation and integration), Ohm's Law, Relationship between charge, current and time.
	**Prerequisites**: Definition of current, relationship between power and resistance,  basic calculus.
## SOLUTION PATHWAY
**Given Information**:
	Charge Q = at - bt^2
	Resistance = R
	Positive constants a and b
**Method & Steps**:
	1. **Find the current (I) as a function of time by differentiating the charge (Q) with respect to time (t).**
		Reasoning: Current is the rate of change of charge with respect to time.
		Formula/Rule: I = dQ/dt
		Working: I = d(at - bt^2)/dt = a - 2bt
	2. **Find the time (t) when the current (I) becomes zero.**
		Reasoning: The current will become zero at some point, and this will be the end of the heat generation.
		Working: 0 = a - 2bt => t = a/2b
	3. **Find the instantaneous power (P) dissipated in the resistor.**
		Reasoning: Power is the product of the square of the current and the resistance.
		Formula/Rule: P = I^2R
		Working: P = (a - 2bt)^2R
	4. **Find the total heat (H) generated by integrating the power (P) from t=0 to t=a/2b.**
		Reasoning: Total heat is the integral of power with respect to time.
		Formula/Rule: H = ∫P dt
		Working: H = ∫[0, a/2b] (a - 2bt)^2R dt  = R∫[0, a/2b] (a^2 - 4abt + 4b^2t^2)dt
		H = R[a^2t - 2abt^2 + (4b^2t^3)/3] [0, a/2b]
		H = R[a^2(a/2b) - 2ab(a/2b)^2 + (4b^2(a/2b)^3)/3]
		H = R[a^3/2b - a^3/2b + a^3/6b]
		H = Ra^3/6b
	5. **Final Answer**
		Result: The total heat produced in R is  H = a^3R/6b
		Verification: The units are consistent (energy/heat).
## OPTION ANALYSIS
[A] ✓: This is the correct answer, as calculated above.
[B] :x: Incorrect, the integration and calculation is not correct
[C] :x: Incorrect, the integration and calculation is not correct
[D] :x: Incorrect, the integration and calculation is not correct
## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrect Differentiation/Integration**: Make sure to apply the power rule of calculus correctly.
		2. **Incorrect Time Limit**: The integration limits are not from 0 to infinity but till the time the current becomes zero.
	**Quick Tips**:
		1. **Understanding the Concepts**: It is important to understand the relationships between charge, current, and time.
		2. **Careful Calculation**: Pay attention to the algebraic manipulation and integration steps to avoid errors.

SUBHANI · Answer

The correct Option is "$\frac{{{{\rm{a}}^{\rm{3}}}{\rm{R}}}}{{{\rm{6b}}}}$"

SUBHANI · Answer

$\frac{{{{\rm{a}}^{\rm{3}}}{\rm{R}}}}{{{\rm{3b}}}}$

SUBHANI · Answer

$\frac{{{{\rm{a}}^{\rm{3}}}{\rm{R}}}}{{{\rm{2b}}}}$

SUBHANI · Answer

$\frac{{{{\rm{a}}^{\rm{3}}}{\rm{R}}}}{{\rm{b}}}$

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