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.

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 $\text{C. 10}\Omega$.