Final answer:
Option (a), with 5V and 9V batteries and 30Ω and 50Ω resistors in series, would create a circuit with a current of 0.175A, meeting the requirements using Ohm's Law.
Step-by-step explanation:
To determine which combination of batteries and resistors can be used to form a circuit diagram with a current of 0.175A, we apply Ohm's Law, which states that Voltage (V) equals Current (I) multiplied by Resistance (R), or V = IR. We can calculate the total voltage needed by rearranging the formula to V = I * R. For a current of 0.175A and potential resistor combinations of 10Ω, 20Ω, 30Ω, 40Ω, and 50Ω, we would need a total resistance that would yield the desired current using one of the individual batteries or a combination thereof.
Since we are connecting in series, the total resistance is the sum of individual resistors: Rtotal = R1 + R2 + R3 ... + Rn. If we choose to use the 9V and 5V batteries in combination, we get a total voltage of 14V. Now, we use Ohm's Law to find the necessary total resistance: Rtotal = V / I = 14V / 0.175A = 80Ω. By selecting the 30Ω and 50Ω resistors and adding them in series (30Ω + 50Ω = 80Ω), we can achieve the desired current of 0.175A when using the 5V and 9V batteries in series.
Therefore, option (a), which consists of using the 5V and 9V batteries and the 30Ω and 50Ω resistors connected in series, would match the requirements for a current of 0.175A. Options (b) and (c) include different combinations that do not meet the resistance requirement for the given current when the batteries they include are used in series, based on the Ohm's Law calculation.