In a series circuit, the total resistance is equal to the sum of the resistances of the individual resistors. The given ratio corresponds to Option D) R1=1, R2=-3, R3=-2, R4=5.
In a series circuit, the total resistance is equal to the sum of the resistances of the individual resistors. Let's assume that the resistance of one of the resistors is x. According to the given ratio, this resistor's resistance divided by the total resistance (-3/5) is equal to x / (R1 + R2 + R3 + R4) = -3/5. To find the set of resistors that corresponds to this ratio, we can substitute the given resistance values and solve for x:
x / (0 + 5 + 7.5 + 11) = -3/5
Simplifying the equation, we get:
x = -(3/5) * (0 + 5 + 7.5 + 11)
After evaluating the expression, we find that x = -22.2 Ω. Therefore, this ratio corresponds to Option D) R1=1, R2=-3, R3=-2, R4=5.
The probable question may be:
In the context of a circuit analysis, the parameter -3/5 represents a specific quantity related to electrical components. Considering a resistor network with different resistances, if the ratio of one resistor's value to the total resistance is -3/5, which set(s) of resistors might this ratio correspond to? Assume the total resistance is represented by the sum of the resistances in the circuit.
Additional Information:
Consider a circuit with resistors R1, R2, R3, and R4.
Total Resistance (R_total) = R1 + R2 + R3 + R4
The ratio of one resistor's resistance to the total resistance is expressed as -3/5.
Options:
A) R1=−3,R2=2,R3=2,R4=4R1=−3,R2=2,R3=2,R4=4
B) R1=6,R2=−10,R3=6,R4=3R1=6,R2=−10,R3=6,R4=3
C) R1=2,R2=5,R3=−3,R4=6R1=2,R2=5,R3=−3,R4=6
D) R1=1,R2=−3,R3=−2,R4=5R1=1,R2=−3,R3=−2,R4=5