(c) hala 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 bulb becomes less bright. what has happened to the ammeter reading?
1
(d) hala replaces the bulb in circuit y to make circuit z. he notices that the ammeter reading is now 2.0 a.
(i) optional for g2 science
describe how the brightness of the bulb changes.
1
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.

Question

(c) hala 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 bulb becomes less bright. what has happened to the ammeter reading?
1
(d) hala replaces the bulb in circuit y to make circuit z. he notices that the ammeter reading is now 2.0 a.
(i) optional for g2 science
describe how the brightness of the bulb changes.
1
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.

Accepted Answer

### Part (c) # Brief Explanations: When the bulb becomes less bright, it means the power $ P = VI $ (or $ I^2R $) has decreased. Assuming the voltage source is constant (e.g., a battery with relatively constant voltage), the brightness is related to the current (and resistance, via $ P = I^2R $ or $ P=\frac{V^2}{R} $). If the bulb is less bright, the current through it has decreased (since $ P \propto I $ for a given $ R $ or $ V $). So the ammeter reading (which measures current) has decreased (become less than 1.0 A). # Answer: The ammeter reading has decreased (is less than 1.0 A). ### Part (d)(i) # Brief Explanations: Brightness of a bulb is related to the power it dissipates, $ P = VI $ (or $ I^2R $, $ \frac{V^2}{R} $). The ammeter reading (current $ I $) has increased from (presumably) the original in circuit Y to 2.0 A in circuit Z. Assuming the voltage $ V $ across the bulb is constant (same circuit voltage source), a higher current means higher power ($ P \propto I $ for constant $ V $ and $ R $, or $ P = I^2R $ if $ R $ is constant). So the bulb becomes brighter. # Answer: The bulb becomes brighter. ### Part (d)(ii) # Brief Explanations: Electrical resistance is a property of a material (or component like a bulb) that opposes the flow of electric current. It is defined as the ratio of the voltage ($ V $) across the component to the current ($ I $) flowing through it, $ R=\frac{V}{I} $. It quantifies how much the component resists the movement of charge carriers (electrons) in the circuit. # Answer: Electrical resistance is the property of a component (or material) that opposes the flow of electric current, defined as $ R = \frac{V}{I} $ (voltage across it divided by current through it). ### Part (d)(iii) # Explanation: ## Step 1: Recall Ohm's Law and Power - Current Relationship For circuit Y, let the voltage be $ V $ (assuming the same voltage source for circuit Z, as it's a replacement in the same circuit setup). In circuit Y, resistance $ R_Y = 2\ \Omega $, and current $ I_Y $ (original in Y before replacement, but in Z, $ I_Z = 2.0\ A $). From Ohm's Law, $ V = I_Y R_Y $ (for Y) and $ V = I_Z R_Z $ (for Z). ## Step 2: Analyze Resistance Change If the current in Z ($ I_Z = 2.0\ A $) is higher than the current in Y (let's say $ I_Y $ was, for example, less than 2.0 A, since in Z it's 2.0 A), and $ V $ is constant (same circuit voltage), then from $ R=\frac{V}{I} $, a higher current $ I $ means a lower resistance $ R $ (because $ R \propto \frac{1}{I} $ for constant $ V $). Alternatively, power $ P = I^2R $; if the bulb is brighter (from (i)), higher current and higher power could imply, but more directly, using $ R=\frac{V}{I} $: if $ I $ increases, $ R $ decreases (since $ V $ is constant). Wait, but wait—if the bulb is brighter, its temperature increases, and the resistance of a bulb (filament, which is a metal) increases with temperature (because metals have positive temperature coefficient of resistance). Wait, maybe I made a mistake earlier. Let's correct: Wait, when a bulb is brighter, it's hotter. The filament's resistance increases with temperature (since for metals, $ R = R_0(1 + \alpha \Delta T) $, where $ \alpha>0 $ for metals). So in circuit Z, current is 2.0 A (higher than, say, the current in Y before replacement). Let's assume in circuit Y, before replacement, the current was $ I_Y $, and in Z, $ I_Z = 2.0\ A $. If the bulb is brighter (hotter) in Z, its resistance $ R_Z $ is higher than $ R_Y $ (because temperature increased). But wait, from Ohm's Law, $ R=\frac{V}{I} $. If $ I_Z > I_Y $, and $ V $ is constant, then $ R_Z=\frac{V}{I_Z} $, $ R_Y=\frac{V}{I_Y} $. If $ I_Z > I_Y $, then $ R_Z < R_Y $ only if $ V $ is constant and we ignore temperature. But the temperature effect: when the bulb is on, its resistance increases with temperature. So if the current is higher, the bulb is hotter, so its resistance is higher than when it was at a lower current (and cooler). Wait, let's clarify: Case 1: Same voltage source (e.g., battery). In circuit Y, bulb has $ R_Y = 2\ \Omega $, current $ I_Y=\frac{V}{2} $. In circuit Z, after replacement, current $ I_Z = 2.0\ A $, so $ R_Z=\frac{V}{2.0\ A} $. If $ I_Z = 2.0\ A $ is higher than $ I_Y $ (e.g., if in Y, $ I_Y = 1.0\ A $, then $ V = 1.0\ A\times2\ \Omega = 2\ V $, so in Z, $ R_Z=\frac{2\ V}{2.0\ A}=1\ \Omega $, which is less than 2\ \Omega. But this ignores temperature. However, when the bulb is brighter (higher current), it's hotter, so its resistance should be higher than when it's cooler (lower current). Wait, there's a contradiction here. The key is: when we replace the bulb, are we replacing it with a different bulb, or is the same bulb operating at higher current? The question says "replaces the bulb", so it's a different bulb. So the new bulb in circuit Z has a current of 2.0 A. Let's use Ohm's Law: $ R=\frac{V}{I} $. If the voltage $ V $ across the bulb is the same as in circuit Y (same circuit, so same $ V $), then for circuit Y, $ V = I_Y \times 2\ \Omega $. For circuit Z, $ V = 2.0\ A \times R_Z $. If $ I_Y $ was, say, 1.0 A (from part (c)), then $ V = 1.0\ A\times2\ \Omega = 2\ V $. Then in circuit Z, $ R_Z=\frac{2\ V}{2.0\ A}=1\ \Omega $, which is less than 2\ \Omega. But if the bulb is brighter, its own resistance (due to temperature) would be higher, but since we replaced the bulb, it's a new bulb. Wait, maybe the question assumes the same voltage, and the replacement is a bulb with lower resistance, so current increases. So the resistance of the bulb in circuit Z is smaller than 2 ohms (because $ R=\frac{V}{I} $, higher $ I $ means lower $ R $ for same $ V $). ## Step 2: Conclusion on Resistance Since $ I_Z = 2.0\ A $ is higher than the current $ I_Y $ in circuit Y (where $ R_Y = 2\ \Omega $, so $ I_Y=\frac{V}{2\ \Omega} $), and $ V $ is constant (same circuit voltage), from $ R=\frac{V}{I} $, a higher current $ I $ implies a lower resistance $ R $. So $ R_Z < 2\ \Omega $. # Answer: The resistance of the bulb in circuit Z is smaller than 2 ohms.

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Question

Explanation:

Part (c)

Answer:

Part (d)(i)