Final answer:
To achieve a current of 1.2 amperes in the circuit with a 40 ohm resistor in series with a 100 ohm resistor (resistor A) and an unknown resistor (resistor B) in parallel, resistor B must be adjusted to have a resistance of 150 ohms.
Step-by-step explanation:
To determine the resistance to which resistor B must be adjusted to achieve a current of 1.2 amperes through the entire circuit, we'll use Ohm's law and the formula for parallel and series resistances.
First, we calculate the total resistance needed for the entire circuit to allow a current of 1.2 A when the supply voltage is 120 V:
Using Ohm's law, V = I * R, we get
R = V / I
= 120 V / 1.2 A
= 100 Ω.
Now, we know that the series resistance (the 40 Ω resistor) must be added to the combined resistance of resistors A and B in parallel to get the total resistance of 100 Ω. Since the resistance of A is 100 Ω and R_series is 40 Ω:
R_total_parallel = R_total - R_series
= 100 Ω - 40 Ω
= 60 Ω
The equivalent resistance for resistors A and B in parallel (R_A and R_B) can be found with 1/R_total_parallel = 1/R_A + 1/R_B. Since 1/R_A = 1/100, we can solve for 1/R_B:
1/R_B = 1/R_total_parallel - 1/R_A
1/R_B = 1/60 Ω - 1/100 Ω
1/R_B = 5/300 Ω - 3/300 Ω
= 2/300 Ω
R_B = 300 Ω / 2
= 150 Ω
Therefore, resistor B must be adjusted to a resistance of 150 Ω to allow the desired 1.2 A of current through the entire circuit. This corresponds to option (d).