Question

Many resistors are connected purely in parallel to a 19 battery and the power output of the battery is 197 Watts. One resistor is removed (but the rest remain connected in parallel to the same battery) and the current coming out of the battery is 6.2 amps with that resistor removed. What was the value of the resistance of the removed resistor in Ohms?

Answer 1

Let the resistance removed in the from the configuration be r and the rest of the resistances in parallel have a net resistance of R.

In the first case, the total resistance is due to R and r in parallel. Thus, R_net = 1/[(1/r) + (1/R)].

Now, power P = V²/R_net , where P = 197W and V = 19V.

⇒ 197 = 19²/{1/[(1/r) + (1/R)]} = 19²[(1/r) + (1/R)] ⇒ 197/361 = (1/r) + (1/R)

⇒ 0.54 = (1/r) + (1/R) ----------------------(1)

In the second case when r is removed, we are given that the battery is the same and the current through the net resistance R is 6.2A. Thus, using V = IR:

19 = 6.2R ⇒ 1/R = 6.2/19 = 0.32 ----------------------(2)

Substituting (2) in (1):

0.54 = (1/r)+0.32 ⇒ (1/r) = 0.54 - 0.32 = 0.22 ⇒ r = 1/0.22 = 4.54 Ω

The value of the resistance removed is 4.54 Ω.