To solve this problem, we need to analyze the circuit and apply the principles of Kirchhoff's laws.
Let's denote the resistances as follows:
- \( R_1 \) for the resistor connected to voltmeter 1,
- \( R_2 \) for the resistor connected to voltmeter 2, and
- \( R_x \) for the unknown resistor.
Given:
- Voltage reading on voltmeter 1 (\( V_1 \)) = 30 V
- Voltage reading on voltmeter 2 (\( V_2 \)) = 10 V
Since the voltmeters are connected in parallel, the voltage across them should be the same. So, we can write:
\[
V_1 = V_2 = V
\]
Now, let's apply Ohm's law to calculate the current (\( I \)) passing through each resistor:
\[
V = I \cdot R
\]
For \( R_1 \):
\[
30 = I \cdot R_1 \quad \text{(1)}
\]
For \( R_2 \):
\[
10 = I \cdot R_2 \quad \text{(2)}
\]
Since \( R_1 \) and \( R_2 \) are in series, the total resistance (\( R_{total} \)) in that branch is \( R_1 + R_2 \).
Given that the voltage (\( V \)) across \( R_x \) is the same as across \( R_{total} \) and applying Ohm's law, we have:
\[
V = I \cdot R_{total} = I \cdot (R_1 + R_2) = I \cdot (30 + 10) = I \cdot 40
\]
Now, let's solve equations (1) and (2) for \( I \):
From (1):
\[
I = \frac{30}{R_1}
\]
From (2):
\[
I = \frac{10}{R_2}
\]
Since \( I \) is the same in both equations, we can equate the expressions for \( I \):
\[
\frac{30}{R_1} = \frac{10}{R_2}
\]
Cross-multiplying, we get:
\[
30 \cdot R_2 = 10 \cdot R_1
\]
Dividing both sides by 10, we get:
\[
3 \cdot R_2 = R_1
\]
Now, we know that \( I = \frac{30}{R_1} \), so substituting \( R_1 = 3 \cdot R_2 \), we get:
\[
I = \frac{30}{3 \cdot R_2} = \frac{10}{R_2}
\]
Since \( I = \frac{V}{R_{total}} = \frac{V}{40} \), we have:
\[
\frac{10}{R_2} = \frac{V}{40}
\]
Solving for \( R_2 \):
\[
R_2 = \frac{40}{10} = 4 \Omega
\]
Now, we can find \( R_1 \):
\[
R_1 = 3 \cdot R_2 = 3 \cdot 4 = 12 \Omega
\]
And finally, the current (\( I \)) passing through the circuit:
\[
I = \frac{V}{R_{total}} = \frac{30}{12 + 4} = \frac{30}{16} = 1.875 A
\]
So, the reading on the ammeter is \( 1.875 \, \text{A} \) and the value of resistor \( R_x \) is \( 4 \, \Omega \).
