When the switch in the circuit in Figure 5 is closed the reading on voltmeter 1 is 30 V and that on voltmeter 2 is 10 V. Determine the reading on the ammeter and the value of resistor Rx​

Respuesta :

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 \( 4 \, \Omega \).

ACCESS MORE
EDU ACCESS