Here is the step-by-step solution to the problems presented in the worksheet.
Part A: Circuit I
Given:
* Voltage of battery ($V$) = $3\text{ V}$
* Resistance of Lamp 1 ($R_1$) = $2\ \Omega$
a) Calculate the current through Lamp L₁.
Using Ohm's Law ($I = \frac{V}{R}$):
$$I_1 = \frac{3\text{ V}}{2\ \Omega} = 1.5\text{ A}$$
b) Calculate the current measured by Ammeter "A".
In Circuit I, there is only one path for the electricity to flow (from the battery, through the lamp, through the ammeter, and back). Therefore, the current is the same everywhere in the circuit.
$$I_{\text{total}} = 1.5\text{ A}$$
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Part B: Circuit II
Given:
* Voltage of battery ($V$) = $3\text{ V}$
* Resistance of Lamp 1 ($R_1$) = $2\ \Omega$
* Resistance of Lamp 2 ($R_2$) = $1\ \Omega$
* The lamps are connected in
parallel.
c) Calculate the total resistance of L₁ and L₂.
For resistors in parallel, the formula is:
$$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}$$
Substitute the values:
$$\frac{1}{R_{\text{total}}} = \frac{1}{2} + \frac{1}{1}$$
$$\frac{1}{R_{\text{total}}} = 0.5 + 1 = 1.5$$
Now, solve for $R_{\text{total}}$:
$$R_{\text{total}} = \frac{1}{1.5} = \frac{2}{3}\ \Omega \approx 0.67\ \Omega$$
d) Did Lamp L₁ get brighter or stay the same brightness when Lamp L₂ was added?
In a parallel circuit connected to an ideal battery, each branch gets the full voltage of the battery.
* In Circuit I, Lamp 1 had $3\text{ V}$ across it.
* In Circuit II, Lamp 1 still has $3\text{ V}$ across it because it is connected directly to the battery terminals.
* Since the voltage and resistance for Lamp 1 did not change, its power (brightness) stays the same.
Answer: Stay the same brightness.
e) What is the current through Lamp L₁ in Circuit II?
Since the voltage across Lamp 1 is still $3\text{ V}$ and its resistance is $2\ \Omega$:
$$I_1 = \frac{3\text{ V}}{2\ \Omega} = 1.5\text{ A}$$
f) What is the current through Lamp L₂ in Circuit II?
Lamp 2 also receives the full battery voltage ($3\text{ V}$). Its resistance is $1\ \Omega$.
$$I_2 = \frac{V}{R_2} = \frac{3\text{ V}}{1\ \Omega} = 3.0\text{ A}$$
g) Calculate the current measured by Ammeter "A" in Circuit II.
The ammeter is placed on the main line next to the battery, so it measures the
total current. In a parallel circuit, the total current is the sum of the currents in each branch.
$$I_{\text{total}} = I_1 + I_2$$
$$I_{\text{total}} = 1.5\text{ A} + 3.0\text{ A} = 4.5\text{ A}$$
*(Self-Check: We can also verify this using the total resistance calculated in step c. $I = \frac{V}{R_{\text{total}}} = \frac{3}{2/3} = 3 \times \frac{3}{2} = 4.5\text{ A}$. The calculations match.)*
Final Answer:
a) 1.5 A
b) 1.5 A
c) 0.67 Ω (or 2/3 Ω)
d) Stay the same brightness
e) 1.5 A
f) 3.0 A
g) 4.5 A
Parent Tip: Review the logic above to help your child master the concept of ohms law practice worksheet.
