Physics Solution Q29

Solution to Question 29

Step 1: Understand the Ohmmeter Reading Definition

The problem states that the ohmmeter is calibrated to read the ratio of the potential difference ($V$) across its terminals to the current ($I$) flowing through it. $$ R_{\text{reading}} = \frac{V_{\text{terminal}}}{I_{\text{branch}}} $$ This means the current through an ohmmeter reading $R_x$ is $I = \frac{V}{R_x}$.

Step 2: Apply KCL at the Common Node

The circuit consists of two ohmmeters and one unknown resistor $R$ all connected in parallel.
Let $V$ be the potential difference across the parallel combination.
Let $I_1$ be the current flowing out of Ohmmeter 1.
Let $I_2$ be the current flowing out of Ohmmeter 2.
Let $I_R$ be the current flowing through the resistor $R$.

According to Kirchhoff’s Current Law (KCL), the sum of currents entering the node from the active sources (ohmmeters) must equal the current leaving through the passive load ($R$): $$ I_1 + I_2 = I_R $$

Step 3: Substitute Resistance Relationships

Using the calibration definition: $$ I_1 = \frac{V}{R_1} \quad \text{and} \quad I_2 = \frac{V}{R_2} $$ For the resistor $R$, by Ohm’s Law: $$ I_R = \frac{V}{R} $$ Substituting these into the KCL equation: $$ \frac{V}{R_1} + \frac{V}{R_2} = \frac{V}{R} $$

Step 4: Solve for R

Dividing by $V$: $$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R} $$ $$ R = \frac{R_1 R_2}{R_1 + R_2} $$ Given $R_1 = 30 \, \Omega$ and $R_2 = 60 \, \Omega$: $$ R = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20 \, \Omega $$

Final Answer

The true value of the unknown resistance is 20 Ω.