By Kirchoffs voltage law the sum of the voltage in the circuit. Consider the circuit with variable resistor R.

5V 10ohms

R
A. find a relationship between I and R in the circuit using E=I R and kirchoffs voltage law.
B. the power to a resistor is given by I^2R watts write P as a function of R.
C. Find the setting of the variable resistor R so that it takes max power

A. To find the relationship between current (I) and variable resistor (R) in the circuit using Kirchhoff's voltage law, we need to analyze the circuit.

Let's assume that the voltage across the 10-ohm fixed resistor is V1 and the voltage across the variable resistor R is V2. According to Kirchhoff's voltage law, the sum of the voltages in a closed loop is zero.

Applying Kirchhoff's voltage law to the circuit:

5V - V1 - V2 = 0

Since the voltage across a resistor is given by Ohm's law (V = I * R), we can rewrite the equation:

5V - (I * 10Ω) - (I * R) = 0

Simplifying the equation:

5V - 10ΩI - RI = 0

Now, let's rearrange the terms:

RI + 10ΩI = 5V

Factoring out the current I:

I(R + 10Ω) = 5V

Finally, we get the relationship between current (I) and variable resistor (R):

I = 5V / (R + 10Ω)

B. The power (P) to a resistor is calculated using the formula P = I^2 * R, where I is the current passing through the resistor and R is the resistance.

Substituting the relationship we found in part A, we can express the current (I) in terms of R:

I = 5V / (R + 10Ω)

Now, we can substitute the expression for I in the power formula:

P = (5V / (R + 10Ω))^2 * R

Simplifying the equation:

P = (25V^2 / (R + 10Ω)^2) * R

Therefore, the power (P) as a function of R is:

P = (25VR^2) / (R + 10Ω)^2

C. To find the setting of the variable resistor R that maximizes the power, we need to find the value of R that gives the maximum value of the function P = (25VR^2) / (R + 10Ω)^2.

To maximize the power, we can differentiate the power function with respect to R and set the derivative equal to zero. Taking the derivative:

dP/dR = (50VR(R + 10Ω)^2 - 50VR^2(2R + 20Ω)) / (R + 10Ω)^4

Setting the derivative equal to zero:

(50VR(R + 10Ω)^2 - 50VR^2(2R + 20Ω)) / (R + 10Ω)^4 = 0

Simplifying the equation:

50VR(R + 10Ω)^2 - 50VR^2(2R + 20Ω) = 0

Dividing both sides by 50VR:

(R + 10Ω)^2 - R(2R + 20Ω) = 0

Expanding and simplifying:

R^2 + 20R + 100Ω^2 - 2R^2 - 20R = 0

Rearranging the terms:

-R^2 + 100Ω^2 = 0

Simplifying further:

R^2 = 100Ω^2

Taking the square root, we get two solutions:

R = 10Ω (positive square root)
R = -10Ω (negative square root)

Since resistance cannot be negative, the setting of the variable resistor R that maximizes the power is R = 10Ω.