Question

Два резистора сопротивлением R1=3,0 Oм и R2=6,0 Oм включены в цепь параллельно. В первом резисторе проходит ток I1=2,0 А. Найдите количество теплоты Q которое выделится обоими резисторами за время t= 10c

Answer 1

Q=Iобщ^2*Rобщ*t - закон Джоуля-Ленца

Соединение параллельное, поэтому общее сопротивление равно:
1/Rобщ=1/R1 + 1/R2
1/Rобщ=1/3 + 1/6 
Rобщ=2 Ом

Iобщ = I1+ I2

Uобщ=U1=U2...

U1=I1*R1= 2*3= 6 B = U2

I2 = U2/R2 = 6/6 = 1 A

Iобщ = 2+1=3 A

Q=3^2 * 2 * 10 = 180 Дж

Ответ: Q = 180 Дж

Answer 2

Calculation of Heat Generated by Resistors in Parallel

To calculate the amount of heat generated by the resistors in parallel, we need to use the formula for calculating the heat generated by a resistor:

Q = I^2 * R * t

Where: - Q is the heat generated (in joules) - I is the current passing through the resistor (in amperes) - R is the resistance of the resistor (in ohms) - t is the time for which the current flows (in seconds)

In this case, we have two resistors, R1 and R2, connected in parallel. The current passing through R1 is given as I1 = 2.0 A. The resistance of R1 is R1 = 3.0 Ω, and the resistance of R2 is R2 = 6.0 Ω. The time for which the current flows is t = 10 s.

To find the total heat generated by both resistors, we can calculate the heat generated by each resistor separately and then sum them up.

Calculation for R1:

Using the formula Q = I^2 * R * t, we can substitute the values for R1, I1, and t:

Q1 = (2.0 A)^2 * 3.0 Ω * 10 s

Simplifying the equation, we get:

Q1 = 4.0 * 3.0 * 10 J

Calculation for R2:

Using the same formula, we can substitute the values for R2, I1, and t:

Q2 = (2.0 A)^2 * 6.0 Ω * 10 s

Simplifying the equation, we get:

Q2 = 4.0 * 6.0 * 10 J

Total Heat Generated:

To find the total heat generated by both resistors, we sum up the individual heat values:

Q = Q1 + Q2

Substituting the values, we get:

Q = (4.0 * 3.0 * 10 J) + (4.0 * 6.0 * 10 J)

Simplifying the equation, we get:

Q = 120 J + 240 J

Therefore, the total amount of heat generated by both resistors in 10 seconds is:

Q = 360 J.

Please note that the above calculations assume that the resistors are ideal and do not take into account any other factors such as temperature coefficients or power dissipation limits.