A47 kΩresistorand 39 kΩresistor are in parallel, and the pair in series with a 22 kΩresistor. Whatis the resistanceof the combination

Answer 1

To find the equivalent resistance of resistors in parallel, we use the formula:

1/Req = 1/R1 + 1/R2 + 1/R3 + ...

So in this case, the equivalent resistance of the 47 kΩ and 39 kΩ resistors in parallel is:

1/Req = 1/47k + 1/39k
= (39 + 47)/(47 * 39)
= 86/1833
≈ 0.047 kΩ

Next, we can consider the resistance of the pair (0.047 kΩ) in series with the 22 kΩ resistor:

Req = 0.047k + 22k
= 22.047 kΩ

Therefore, the resistance of the combination is approximately 22.047 kΩ.