Question

A total charge q passes through a coil of resistance R. Let H be the amount of heat generated in the coil. Choose the correct statement (s).

• A. $H = \frac{2q^2R}{3t_0}$ if current decreases uniformly to zero during a time interval $t_0$
• B. $H = \frac{4q^2R}{3t_0}$ if current decreases uniformly to zero during a time interval $t_0$
• C. $H = \frac{q^2R}{t_0}$ if the current decreases smoothly down to zero halving its values every $t_0$ seconds
• D. $H = \frac{q^2Rln(2)}{2t_0}$ if the current decreases smoothly down to zero halving its value every $t_0$ seconds

Answer 1

Answer: (B), (D)

H = \frac{4q^2R}{3t_0} if current decreases uniformly to zero during a time interval t_0, H = \frac{q^2Rln(2)}{2t_0} if the current decreases smoothly down to zero halving its value every t_0 seconds

Answer 2

Let $i(t)$ be the current flowing through the coil of resistance $R$ at time $t$. The heat generated in the coil is given by $H = \int R i(t)^2 dt$. The total charge that passes through the coil is given by $q = \int i(t) dt$.

Scenario 1: Current decreases uniformly to zero during a time interval $t_0$.

Let the initial current at $t=0$ be $i_0$. Since the current decreases uniformly to zero in time $t_0$, the current as a function of time is given by a linear equation: $i(t) = i_0 \left(1 - \frac{t}{t_0}\right)$ for $0 \le t \le t_0$.

The total charge $q$ is the area under the $i(t)$ vs $t$ graph from $t=0$ to $t=t_0$. This is a triangle with base $t_0$ and height $i_0$. $q = \frac{1}{2} \times t_0 \times i_0$. Thus, the initial current is $i_0 = \frac{2q}{t_0}$. Substituting this into the expression for $i(t)$: $i(t) = \frac{2q}{t_0} \left(1 - \frac{t}{t_0}\right)$. The heat generated is $H = \int_0^{t_0} R i(t)^2 dt$. $H = \int_0^{t_0} R \left[\frac{2q}{t_0} \left(1 - \frac{t}{t_0}\right)\right]^2 dt = R \frac{4q^2}{t_0^2} \int_0^{t_0} \left(1 - \frac{t}{t_0}\right)^2 dt$.

Let $u = 1 - \frac{t}{t_0}$, so $du = -\frac{1}{t_0} dt$, which means $dt = -t_0 du$. When $t=0$, $u=1$. When $t=t_0$, $u=0$. The integral becomes $\int_1^0 u^2 (-t_0 du) = -t_0 \int_1^0 u^2 du = t_0 \int_0^1 u^2 du$. $\int_0^1 u^2 du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3}$. So, $\int_0^{t_0} \left(1 - \frac{t}{t_0}\right)^2 dt = t_0 \times \frac{1}{3} = \frac{t_0}{3}$. $H = R \frac{4q^2}{t_0^2} \times \frac{t_0}{3} = \frac{4q^2 R}{3t_0}$.

Scenario 2: Current decreases smoothly down to zero halving its value every $t_0$ seconds.

This implies an exponential decay of the form $i(t) = i_0 e^{-kt}$. The condition "halving its value every $t_0$ seconds" means $i(t+t_0) = \frac{1}{2} i(t)$. $i_0 e^{-k(t+t_0)} = \frac{1}{2} i_0 e^{-kt}$ $e^{-kt} e^{-kt_0} = \frac{1}{2} e^{-kt}$ $e^{-kt_0} = \frac{1}{2}$ $-kt_0 = \ln\left(\frac{1}{2}\right) = -\ln(2)$ $k = \frac{\ln(2)}{t_0}$. The current is $i(t) = i_0 e^{-\frac{\ln(2)}{t_0} t}$. The current approaches zero as $t \to \infty$. The total charge $q$ is the integral of the current from $t=0$ to $t=\infty$. $q = \int_0^{\infty} i_0 e^{-\frac{\ln(2)}{t_0} t} dt = i_0 \int_0^{\infty} e^{-\frac{\ln(2)}{t_0} t} dt$. Let $\alpha = \frac{\ln(2)}{t_0}$. The integral is $\int_0^{\infty} e^{-\alpha t} dt = \left[-\frac{1}{\alpha} e^{-\alpha t}\right]_0^{\infty} = 0 - \left(-\frac{1}{\alpha} \times 1\right) = \frac{1}{\alpha}$. $q = i_0 \times \frac{1}{\frac{\ln(2)}{t_0}} = i_0 \frac{t_0}{\ln(2)}$. So, the initial current is $i_0 = \frac{q \ln(2)}{t_0}$. The current is $i(t) = \frac{q \ln(2)}{t_0} e^{-\frac{\ln(2)}{t_0} t}$. The heat generated is $H = \int_0^{\infty} R i(t)^2 dt$. $H = \int_0^{\infty} R \left[\frac{q \ln(2)}{t_0} e^{-\frac{\ln(2)}{t_0} t}\right]^2 dt = R \frac{q^2 (\ln(2))^2}{t_0^2} \int_0^{\infty} e^{-2\frac{\ln(2)}{t_0} t} dt$. Let $\beta = 2\frac{\ln(2)}{t_0}$. The integral is $\int_0^{\infty} e^{-\beta t} dt = \frac{1}{\beta}$. $\int_0^{\infty} e^{-2\frac{\ln(2)}{t_0} t} dt = \frac{1}{2\frac{\ln(2)}{t_0}} = \frac{t_0}{2\ln(2)}$. $H = R \frac{q^2 (\ln(2))^2}{t_0^2} \times \frac{t_0}{2\ln(2)} = \frac{q^2 (\ln(2)) R}{2t_0}$.

The correct statements are the ones corresponding to option 2 and option 4.