Question

Calculate the jul(i)an heat released in 20s on the resistors R1 = 6 ohm and R2 = 3 ohm which are connected in parallel, if the voltage at the ends of the resistor is U = 24V ?​

Answer 1

The heat released in 20s is 5760J

Explanation:

Given:

Time, t = 20s

Resistor, R₁ = 6Ω

               R₂ = 3Ω

Voltage, V = 24V

Heat, H = ?

When the two resistors are connected in parallel, the net resistance would be:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

Substituting the value in the above formula we get:

$\frac{1}{R} = \frac{1}{6} + \frac{1}{3} \\\\\frac{1}{R} = \frac{1+2}{6}$

$\frac{1}{R} = \frac{3}{6} \\\\R = 2$

The net resistance is 2Ω

According to the joule's law:

$H = \frac{V^2}{R} t$

On substituting the value we get:

$H = \frac{(24)^2}{2} X 20\\\\H = 5760J$

Therefore, heat released in 20s is 5760J