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

Step-by-step 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:

`(1)/(R) = (1)/(R_1) + (1)/(R_2)`

Substituting the value in the above formula we get:

`(1)/(R) = (1)/(6) + (1)/(3) \\\\(1)/(R) = (1+2)/(6) \\`

`(1)/(R) = (3)/(6) \\\\R = 2`

The net resistance is 2Ω

According to the joule's law:

`H = (V^2)/(R) t`

On substituting the value we get:

`H = ((24)^2)/(2) X 20\\\\H = 5760J`

Therefore, heat released in 20s is 5760J