Answer:
a. To find the current in the circuit, we can use Ohm's Law which states that current (I) equals voltage (V) divided by resistance (R). The potential difference across R1 is given as 24 V, and we know the voltage of the battery is 40 V. Therefore, the potential difference across the other two resistances in the circuit must be 40 V - 24 V = 16 V.
Using Ohm's Law for each resistance:
I₁ = V₁ / R₁ = 24 V / 240 Ω = 0.1 A
I₂ = V₂ / R₂
I₃ = V₃ / R₃ = 16 V / 120 Ω = 0.133 A
Since the resistances are in series, the same current flows through all of them. Therefore, the current in the circuit is:
I = I₁ = I₂ = I₃ = 0.1 A
b. To find the equivalent resistance of the circuit, we can use the formula for the total resistance of resistors in series, which states that the equivalent resistance (R_eq) is equal to the sum of the individual resistances (R1 + R2 + R3):
R_eq = R1 + R2 + R3
R_eq = 240 Ω + R2 + 120 Ω (since R3 is given as 120 2 which we assume is a typo)
R_eq = 360 Ω + R2
c. To find the resistance of R2, we can rearrange the equation for R_eq:
R_eq = 360 Ω + R2
R2 = R_eq - 360 Ω
We don't know the value of R_eq yet, but we can find it using Ohm's Law again. Since the current is the same through all the resistances in the circuit, we can add up the voltage drops across each resistance to get the total voltage drop across the circuit:
V_total = V1 + V2 + V3
V_total = IR1 + IR2 + IR3
V_total = I(R1 + R2 + R3)
V_total = I*R_eq
We know that V_total is equal to the voltage of the battery, which is 40 V. Therefore:
40 V = I*R_eq
R_eq = 40 V / I
R_eq = 40 V / 0.1 A
R_eq = 400 Ω
Now we can substitute this value into the equation for R2:
R2 = R_eq - 360 Ω
R2 = 400 Ω - 360 Ω
R2 = 40 Ω
Therefore, the resistance of R2 is 40 Ω.