Question

Currents flowing through two conductors of equal resistance produce 16 times more heat in second conductor compared to the first one in a particular time interval. The ratio of currents flowing through the first and second conductors is

• A. 1:4
• B. 4:1
• C. 1:16
• D. 16:1

Answer 1

Answer: (A)

1:4

Answer 2

The heat produced in a conductor due to the flow of current is given by Joule's law of heating:

$H = I^2 R t$

where:

• $H$ is the heat produced
• $I$ is the current flowing through the conductor
• $R$ is the resistance of the conductor
• $t$ is the time for which the current flows

Given:

1. The resistances of the two conductors are equal: $R_1 = R_2 = R$.
2. The heat produced in the second conductor is 16 times more than in the first one: $H_2 = 16 H_1$.
3. The time interval is the same for both conductors: $t_1 = t_2 = t$.

Let $I_1$ be the current in the first conductor and $I_2$ be the current in the second conductor.

For the first conductor:

$H_1 = I_1^2 R_1 t_1$

$H_1 = I_1^2 R t$ (Equation 1)

For the second conductor:

$H_2 = I_2^2 R_2 t_2$

$H_2 = I_2^2 R t$ (Equation 2)

Now, substitute the given relationship $H_2 = 16 H_1$ into the equations:

$I_2^2 R t = 16 (I_1^2 R t)$

Since $R$ and $t$ are common and non-zero on both sides, they can be cancelled out:

$I_2^2 = 16 I_1^2$

To find the ratio of currents $I_1 : I_2$, take the square root of both sides:

$\sqrt{I_2^2} = \sqrt{16 I_1^2}$

$I_2 = 4 I_1$

Now, express this as a ratio $I_1 : I_2$:

$\frac{I_1}{I_2} = \frac{1}{4}$

So, the ratio of currents flowing through the first and second conductors is 1:4.