Question

Draw a circuit diagram showing the above two resistors X and Y joined in parallel with the same battery and same ammeter and voltmeter.

Answer 1

Parallel Resistor Circuit Description

Two resistors: \[ R_X = 3\,\Omega \quad \text{and} \quad R_Y = 6\,\Omega \] are connected in parallel to a \[ 2\,\text{V battery} \] with an ammeter measuring total current and a voltmeter across the parallel combination.

Circuit Diagram Representation (ASCII Style)

        + 2V -         ┌────┬────────────┬───────┐         │    │            │       │         │   [3Ω]         [6Ω]     │         │    │            │       │        (V)  └────┬────────┘       │                 │                │                (A)              GND  

Key Observations:

• Resistors X (3 Ω) and Y (6 Ω) are in parallel
• Voltage across both resistors is the same: \( V = 2\,\text{V} \)
• Current through each resistor:
  • \( I_X = \frac{V}{R_X} = \frac{2}{3} \, \text{A} \)
  • \( I_Y = \frac{V}{R_Y} = \frac{2}{6}\)

Answer 2

In the case of parallel combination:
- Potential Difference Across X and Y: In parallel, the potential difference across both resistors X and Y will be the same because both resistors are connected across the same battery. Therefore, the potential difference across X and Y is the same in a parallel combination. - Current Through X and Y: In a parallel combination, the total current supplied by the battery divides between the two resistors. However, the current through each resistor will not be the same unless their resistances are equal. In the case of series combination:
- The same current flows through both X and Y as they are connected in series.
- However, the potential difference across each resistor will be different due to the difference in their resistances.

Answer 3

To find the current drawn from the battery when the resistors are connected in series, we first calculate the total resistance of the series combination. \[ R_{\text{total}} = R_X + R_Y = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega \] Now, we use Ohm's law to find the current drawn from the battery: \[ I = \frac{V}{R_{\text{total}}} = \frac{2 \, \text{V}}{9 \, \Omega} = 0.222 \, \text{A} \] Therefore, the current drawn from the battery by the series combination is 0.222 A.

Answer 4

To calculate the equivalent resistance of the parallel combination of resistors X and Y, we use the formula for parallel resistors: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_X} + \frac{1}{R_Y} \] Substituting the given values: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{3 \, \Omega} + \frac{1}{6 \, \Omega} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \] Thus, the equivalent resistance is: \[ R_{\text{eq}} = 2 \, \Omega \]