Question

Resistors of 20 ohms, 20 ohms, and 30 ohms are connected in parallel. What resistance must be added in series with the combination to obtain a total resistance of 10 ohms. If the complete circuit expends 0.36 kW, find the total current flowing.​

Answer 1

Given:

R1 = 20 ohms

R2 = 20 ohms

R3 = 30 ohms

Total resistance = 10 ohms

Power = 0.36 kW

1/Req = 1/20 + 1/20 + 1/30

1/Req=2/15

Req 15/2

Req = 7.5 ohms

Hence, 10 - 7.5 = 2.5 ohms

PIR, hence 0.36 x 103

= I²(10)

I = √360/10 = √36 = 6 A