Question

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.

Answer 1

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