Question

R1=20 Om, r2= 75 Om соедино паралельно I2=?, P1=0.36wt

Answer 1

Найдем напряжение
U=U1=U2=√P1/R1=√0.36/20=0.134V
ток на втором резисторе
I2=U2/R2=0.134/75=1.79 мАмпер

Answer 2

To calculate the total current (I2) flowing through the parallel combination of resistors R1 and R2, we can use the formula for calculating the total resistance in a parallel circuit. The formula is as follows:

1/R_total = 1/R1 + 1/R2

Where R1 and R2 are the resistances of the individual resistors. Let's substitute the given values:

R1 = 20 Ω R2 = 75 Ω

1/R_total = 1/20 Ω + 1/75 Ω

Now, calculate the sum on the right-hand side:

1/R_total = (75 + 20) / (20 * 75) = 95 / 1500 = 19/300

Now, find the reciprocal of R_total:

R_total = 300 / 19 ≈ 15.79 Ω

Now, we can calculate the total current (I2) using Ohm's Law, where I is the current, V is the voltage, and R is the resistance:

P = V^2 / R

Since the power (P1) and resistance (R_total) are given for the parallel combination, we can rearrange Ohm's Law to solve for the total current (I2):

I2 = √(P1 / R_total)

Now, substitute the given values:

P1 = 0.36 W R_total ≈ 15.79 Ω

I2 = √(0.36 / 15.79) ≈ √(0.0228) ≈ 0.151 A (approximately)

Therefore, the total current (I2) flowing through the parallel combination of resistors is approximately 0.151 amperes.