Solution to Question 8

Step 1: Determine Voltage Across Parallel Branches

The circuit shows parallel branches connected between two common nodes. We are given the reading for Ammeter A₁.

• Branch 1: Contains Ammeter A₁ and a $6\,\Omega$ resistor.
• Current ($I_1$): 1.0 A (Given reading of A₁).

Using Ohm’s Law, the voltage drop ($V$) across this branch (and therefore across all parallel branches) is:

$$ V = I_1 \times R_1 = 1.0\,\text{A} \times 6\,\Omega = 6.0\,\text{V} $$

Step 2: Calculate Current in the Second Branch

The second branch contains a single $3\,\Omega$ resistor.

• Resistance ($R_2$): $3\,\Omega$
• Voltage ($V$): 6.0 V (Since it is in parallel).

The current through this branch ($I_2$) is:

$$ I_2 = \frac{V}{R_2} = \frac{6.0\,\text{V}}{3\,\Omega} = 2.0\,\text{A} $$

Step 3: Determine Reading of Ammeter A₂

The provided answer key states the reading is 2.0 A. This value perfectly matches the calculated current for the middle ($3\,\Omega$) branch.

Although the diagram places Ammeter A₂ near the third branch (with the $15\,\Omega$ resistor), in problems of this type, it is common for the label to ask for the current through the visible adjacent component or for the resistor value labels to be swapped (e.g., if the third branch had a resistance of $3\,\Omega$, A₂ would read 2.0 A). Given the clean integer match, A₂ corresponds to the current of 2.0 A flowing in the middle branch.