Question

(c) hafiz sets up circuit y. the ammeter reading is 1.0 a. he goes for his lunch and when he returns after 2 hours, he notices that the becomes less bright. what has happened to the ammeter reading? (d) hafiz replaces the bulb in circuit y to make circuit z. he notices that the reading is now 2.0 a. (i) optional for g2 science describe how the brightness of the bulb changes. the resistance of the bulb in circuit y is 2 ohms. (ii) what is electrical resistance? (iii) optional for g2 science state whether the resistance of the bulb in circuit z will be smaller than greater than or exactly 2 ohms.

Explanation:

(d)(i)

In a circuit, brightness of a bulb is related to power (\(P = VI\) or \(P=I^2R\)). When the ammeter reading (current \(I\)) increases from 1.0 A to 2.0 A (assuming voltage \(V\) is constant or changes appropriately), power increases. Higher power means more light energy, so the bulb in circuit Z should be brighter than in circuit Y.

Electrical resistance is a property of a material or component that opposes the flow of electric current. It is defined as the ratio of voltage (\(V\)) across the component to the current (\(I\)) flowing through it, \(R=\frac{V}{I}\). It is measured in ohms (\(\Omega\)) and depends on factors like material, length, cross - sectional area, and temperature of the conductor.

We know that \(I=\frac{V}{R}\) (Ohm's law). For circuit Y, \(I_Y = 1.0\space A\) and \(R_Y=2\space\Omega\), so \(V = I_YR_Y=1.0\space A\times2\space\Omega = 2\space V\) (assuming the voltage source is the same for circuit Z, or the circuit is designed such that voltage across the bulb is related to the source). For circuit Z, \(I_Z = 2.0\space A\). If we assume the voltage across the bulb in circuit Z is the same as in circuit Y (or related to the source voltage), from \(R=\frac{V}{I}\), when \(I\) increases, \(R\) decreases (since \(V\) is constant or changes in a way that \(V/I\) decreases). Also, for a bulb (a resistor that heats up), as current increases, the bulb gets hotter, but in this case, since current increased and if we consider the formula \(R = \frac{V}{I}\), with \(I\) doubling (from 1 A to 2 A) and assuming \(V\) is constant (or the change in \(V\) is not enough to counteract the current change), the resistance \(R_Z=\frac{V}{I_Z}\) will be smaller than \(R_Y = 2\space\Omega\) (because \(I_Z>I_Y\) and \(R\propto\frac{1}{I}\) when \(V\) is constant).

Answer:

The bulb in circuit Z is brighter than the bulb in circuit Y. Because the current (ammeter reading) increased from 1.0 A to 2.0 A, and brightness depends on power (\(P = I^2R\) or \(P = VI\)). With higher current (assuming voltage or resistance context), power increases, making the bulb brighter.

(d)(ii)