Physics, asked by darkhaxor, 4 months ago

Three resistors are in parallel. The current in the first resistor is 0.1 A. The power dissipated in the
second is 3 watts. The voltage drop across the third is 100 volts. Determine the ohmic values of
resistors and the total resistance if total current is 0.2 A.

Answers

Answered by safinrahman666
11

Answer:

R1 = 1 k Ohms

R2 = 3.333 k Ohms

RE = 1.428 k Ohms

So, Total R = 5.761 k Ohms

Explanation:

Given :

V/R1 = 0.1 Amps

V²/R2 = 3 W

V across R3 = 100 V

Since the resistors are in parallel, voltage across each resistor will be the same, i.e., V = 100 V

So,

V/R1 = 0.1 Amps = i1

100/R1 = 0.1 Amps

R1 = 100 / 0.1 Ohms = 1 k Ohms

V²/R2 = 3 W

100²/R2 = 3 W

R2 = 100²/3 Ohms = 3.333 k Ohms

Also, (i2)² x R2 = 3 W

(i2)² = (3)/(3.333 x 10³)

i2 = 0.03 Amps

so, i3 = i1 - i2 = 0.1 - 0.03 = 0.07 Amps

so, R3 = V/i3 = 100/0.07 = 1.428 k Ohms

Thus, total resistance R = R1 + R2 + R3 = 5.761 k Ohms

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