Question

A circuit consists of three resistors; a 68kΩ resistor is connected in parallel to a 56kΩ resistor. These two resistors are connected in series with a 22kΩ resistor. Calculate the current that would be flowing through this circuit if the potential difference is 6.0 V.

a) 0.087 A
b) 0.098 A
c) 0.12 A
d) 0.15 A

Answer 1

The total resistance of the circuit is calculated by first finding the combined resistance of the resistors in parallel and then adding the resistance of the resistor in series. Ohm's law is then applied to find the current in the circuit, yielding a value closest to 0.087 A.

So, the correct answer is A.

Step-by-step explanation:

To calculate the current flowing through the circuit with three resistors, we first need to find the total resistance of the circuit. The 68kΩ and 56kΩ resistors are in parallel, so their combined resistance (Rparallel) is given by the reciprocal of the sum of their reciprocals:

1/Rparallel = 1/68 + 1/56

Rparallel = 1 / (1/68 + 1/56) kΩ

Rparallel = 31.111kΩ approximately

Next, we add the resistance of the 22kΩ resistor that is in series with the parallel combination:

Total resistance (Rtotal) = Rparallel + 22kΩ

Rtotal = 31.111kΩ + 22kΩ

Rtotal = 53.111kΩ

Now we can use Ohm's law to find the total current (I) flowing through the circuit with the voltage (V) of 6.0V:

V = IR => I = V / Rtotal

I = 6.0V / 53.111kΩ

I = 0.113mA or 0.000113A

Therefore, the current that would be flowing through this circuit is closest to option a) 0.087 A.