Question

The heat generated in a coil of resistance $R$ due to charge $q$ passing through it is $H_1$ when current in the coil decreases down to zero uniformly in time interval $\Delta t$ and $H_2$ when current in the coil decreases down to zero such that current continuously becomes half after each interval of time $\Delta t$. Then $\frac{H_2}{H_1}$ is

• A. $\frac{ln(8)}{8}$
• B. $\frac{ln(5)}{3}$
• C. $\frac{ln(10)}{5}$
• D. $\frac{ln(6)}{8}$

Answer 1

Answer: (A)

$\frac{ln(8)}{8}$

Answer 2

To determine the ratio $\frac{H_2}{H_1}$, we need to calculate the heat generated in two different scenarios. The heat generated in a coil of resistance $R$ is given by $H = \int I^2 R dt$, and the total charge passing through the coil is $q = \int I dt$.

Scenario 1: $H_1$ - Current decreases uniformly down to zero in time interval $\Delta t$.

Let the initial current at $t=0$ be $I_0$. Since the current decreases uniformly to zero in time $\Delta t$, the current as a function of time is given by: $I(t) = I_0 \left(1 - \frac{t}{\Delta t}\right)$ for $0 \le t \le \Delta t$.

The total charge $q$ is the area under the $I(t)$ vs $t$ graph, which is a triangle: $q = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \Delta t \times I_0$. From this, the initial current $I_0 = \frac{2q}{\Delta t}$.

Now, we calculate $H_1$: $H_1 = \int_0^{\Delta t} R I(t)^2 dt = R \int_0^{\Delta t} \left[ \frac{2q}{\Delta t} \left(1 - \frac{t}{\Delta t}\right) \right]^2 dt$ $H_1 = R \frac{4q^2}{(\Delta t)^2} \int_0^{\Delta t} \left(1 - \frac{t}{\Delta t}\right)^2 dt$.

To solve the integral, let $u = 1 - \frac{t}{\Delta t}$. Then $du = -\frac{1}{\Delta t} dt$, so $dt = -\Delta t du$. When $t=0$, $u=1$. When $t=\Delta t$, $u=0$. The integral becomes: $\int_1^0 u^2 (-\Delta t du) = -\Delta t \int_1^0 u^2 du = \Delta t \int_0^1 u^2 du = \Delta t \left[ \frac{u^3}{3} \right]_0^1 = \Delta t \times \frac{1}{3} = \frac{\Delta t}{3}$.

Substitute this back into the expression for $H_1$: $H_1 = R \frac{4q^2}{(\Delta t)^2} \times \frac{\Delta t}{3} = \frac{4q^2 R}{3\Delta t}$.

Scenario 2: $H_2$ - Current continuously becomes half after each interval of time $\Delta t$.

This implies an exponential decay of the current. Let $I(t) = I'_0 e^{-kt}$. The condition "current continuously becomes half after each interval of time $\Delta t$" means $I(t+\Delta t) = \frac{1}{2} I(t)$. $I'_0 e^{-k(t+\Delta t)} = \frac{1}{2} I'_0 e^{-kt}$ $e^{-k\Delta t} = \frac{1}{2}$ $-k\Delta t = \ln\left(\frac{1}{2}\right) = -\ln(2)$ So, $k = \frac{\ln(2)}{\Delta t}$. The current is $I(t) = I'_0 e^{-\frac{\ln(2)}{\Delta t} t}$. The current approaches zero asymptotically, so the integration time is from $t=0$ to $t=\infty$.

The total charge $q$ is: $q = \int_0^{\infty} I(t) dt = \int_0^{\infty} I'_0 e^{-\frac{\ln(2)}{\Delta t} t} dt$. Let $\alpha = \frac{\ln(2)}{\Delta t}$. The integral is $I'_0 \int_0^{\infty} e^{-\alpha t} dt = I'_0 \left[ -\frac{1}{\alpha} e^{-\alpha t} \right]_0^{\infty} = I'_0 \left(0 - (-\frac{1}{\alpha} \times 1)\right) = \frac{I'_0}{\alpha}$. So, $q = I'_0 \frac{\Delta t}{\ln(2)}$, which gives $I'_0 = \frac{q \ln(2)}{\Delta t}$.

Now, we calculate $H_2$: $H_2 = \int_0^{\infty} R I(t)^2 dt = R \int_0^{\infty} \left[ \frac{q \ln(2)}{\Delta t} e^{-\frac{\ln(2)}{\Delta t} t} \right]^2 dt$ $H_2 = R \frac{q^2 (\ln(2))^2}{(\Delta t)^2} \int_0^{\infty} e^{-2\frac{\ln(2)}{\Delta t} t} dt$.

Let $\beta = 2\frac{\ln(2)}{\Delta t}$. The integral is $\int_0^{\infty} e^{-\beta t} dt = \left[ -\frac{1}{\beta} e^{-\beta t} \right]_0^{\infty} = \frac{1}{\beta}$. So, $\int_0^{\infty} e^{-2\frac{\ln(2)}{\Delta t} t} dt = \frac{1}{2\frac{\ln(2)}{\Delta t}} = \frac{\Delta t}{2\ln(2)}$.

Substitute this back into the expression for $H_2$: $H_2 = R \frac{q^2 (\ln(2))^2}{(\Delta t)^2} \times \frac{\Delta t}{2\ln(2)} = \frac{q^2 (\ln(2)) R}{2\Delta t}$.

Calculate the ratio $\frac{H_2}{H_1}$: $\frac{H_2}{H_1} = \frac{\frac{q^2 (\ln(2)) R}{2\Delta t}}{\frac{4q^2 R}{3\Delta t}}$. Cancel out the common terms $q^2 R$ and $\Delta t$: $\frac{H_2}{H_1} = \frac{\frac{\ln(2)}{2}}{\frac{4}{3}} = \frac{\ln(2)}{2} \times \frac{3}{4} = \frac{3\ln(2)}{8}$.

Using the logarithm property $a \ln(b) = \ln(b^a)$: $3\ln(2) = \ln(2^3) = \ln(8)$. Therefore, $\frac{H_2}{H_1} = \frac{\ln(8)}{8}$.