Here's how to determine the value of X: The circuit shows two resistors, X and R, connected in parallel to a voltage source E. An ammeter measures the total current flowing from the source. The equivalent resistance ($R_{eq}$) for two resistors in parallel is given by: $$R_{eq} = \frac{X \cdot R}{X + R}$$ According to Ohm's Law, the voltage (E) across the parallel combination is the total current (I) multiplied by the equivalent resistance: $$E = I \cdot R_{eq}$$ Since the voltage source E is constant, we can set up equations for the two given scenarios. Scenario 1: Current $I_1 = 3 \, \text{A}$ Resistance $R_1 = 5 \, \Omega$ Equivalent resistance $R_{eq1} = \frac{X \cdot 5}{X + 5}$ Voltage $E = 3 \cdot \left(\frac{5X}{X + 5}\right) = \frac{15X}{X + 5}$ (Equation 1) Scenario 2: Current $I_2 = 6 \, \text{A}$ Resistance $R_2 = 2 \, \Omega$ Equivalent resistance $R_{eq2} = \frac{X \cdot 2}{X + 2}$ Voltage $E = 6 \cdot \left(\frac{2X}{X + 2}\right) = \frac{12X}{X + 2}$ (Equation 2) Now, we equate Equation 1 and Equation 2 since E is constant: $$\frac{15X}{X + 5} = \frac{12X}{X + 2}$$ Since X is a resistance, $X \neq 0$. We can divide both sides by X: $$\frac{15}{X + 5} = \frac{12}{X + 2}$$ Cross-multiply to solve for X: $$15(X + 2) = 12(X + 5)$$ $$15X + 30 = 12X + 60$$ Subtract $12X$ from both sides: $$15X - 12X + 30 = 60$$ $$3X + 30 = 60$$ Subtract $30$ from both sides: $$3X = 60 - 30$$ $$3X = 30$$ Divide by $3$: $$X = \frac{30}{3}$$ $$X = 10 \, \Omega$$ The value of X is $10 \, \Omega$. The final answer is $\boxed{\text{C. 10}\Omega}$.