Question

A 100 V battery is connected to the electric network as shown. If the power consumed in the 2Ω resistor is 200 W, determine the power dissipated in the 5Ω resistor.

Answer 1

We know that Power, P =I² R

⇒ 200 = l² x 2

l² = 200/2 = 100

⇒l= √100 = 10A

∴Current flowing through 2Ω resistor = 10 A

Potential drop across 2Ωresistor, V =IR

= 10 × 2 = 20 V

Equivalent resistance of 30Ω and 6Ω

(30 x 6/30+6) = 180/36 = 5Ω

∴Therefore, potential across parallel combination of 40Ω and 10Ω

=10 x 8 = 80 V

∴ Current through 5Ωresistor, l = 80/10 = 8A

∴Power dissipated in 5Ω resistor = l² R = 8² x 5 = 320 W