Ohms Law and Power Practice Problems by Mrs K Science | TPT - Free Printable

Problem Analysis:

The image contains a series of questions related to Ohm's Law, power calculations, and circuit analysis. Let's solve each problem step by step.

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Question 9:

Statement: A source uses 2 V at 0.5 A of current when it is operated by a 12 V battery. What is the resistance of the source?

#### Solution:
Ohm's Law states:
\[
V = IR
\]
where:
- \( V \) is the voltage across the source,
- \( I \) is the current through the source,
- \( R \) is the resistance of the source.

Given:
- Voltage across the source, \( V = 2 \, \text{V} \),
- Current through the source, \( I = 0.5 \, \text{A} \).

Using Ohm's Law:
\[
R = \frac{V}{I} = \frac{2 \, \text{V}}{0.5 \, \text{A}} = 4 \, \Omega
\]

#### Final Answer:
\[
\boxed{4 \, \Omega}
\]

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Question 10:

Statement: A lamp draws a current of 0.3 A when it is connected to a 12 V source.
1. What is the resistance of the lamp?
2. What is the power consumption of the lamp?

#### Solution:
1. Resistance of the lamp:
Using Ohm's Law:
\[
R = \frac{V}{I}
\]
Given:
- Voltage, \( V = 12 \, \text{V} \),
- Current, \( I = 0.3 \, \text{A} \).

Substituting the values:
\[
R = \frac{12 \, \text{V}}{0.3 \, \text{A}} = 40 \, \Omega
\]

2. Power consumption of the lamp:
The power consumed by the lamp can be calculated using the formula:
\[
P = VI
\]
Substituting the given values:
\[
P = 12 \, \text{V} \times 0.3 \, \text{A} = 3.6 \, \text{W}
\]

#### Final Answers:
1. Resistance of the lamp: \(\boxed{40 \, \Omega}\)
2. Power consumption of the lamp: \(\boxed{3.6 \, \text{W}}\)

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Question 11:

Statement: A lamp is connected to 120 V.
1. What is the resistance of the lamp if its power is 60 W?
2. What is the current in the lamp?

#### Solution:
1. Resistance of the lamp:
The power consumed by the lamp is given by:
\[
P = \frac{V^2}{R}
\]
Rearranging for resistance:
\[
R = \frac{V^2}{P}
\]
Given:
- Voltage, \( V = 120 \, \text{V} \),
- Power, \( P = 60 \, \text{W} \).

Substituting the values:
\[
R = \frac{(120 \, \text{V})^2}{60 \, \text{W}} = \frac{14400 \, \text{V}^2}{60 \, \text{W}} = 240 \, \Omega
\]

2. Current in the lamp:
Using Ohm's Law:
\[
I = \frac{V}{R}
\]
Substituting the values:
\[
I = \frac{120 \, \text{V}}{240 \, \Omega} = 0.5 \, \text{A}
\]

#### Final Answers:
1. Resistance of the lamp: \(\boxed{240 \, \Omega}\)
2. Current in the lamp: \(\boxed{0.5 \, \text{A}}\)

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Question 12:

Statement: A resistor is added in series to a bulb so that the power becomes one-half of its original value.
1. What is the potential difference (voltage) across the bulb?
2. How much resistance was established in the circuit?
3. How much power was lost in the resistor?

#### Solution:
Let the original resistance of the bulb be \( R_b \), and the original voltage across the bulb be \( V \). The original power dissipated by the bulb is:
\[
P_{\text{original}} = \frac{V^2}{R_b}
\]

When a resistor \( R_r \) is added in series, the total resistance becomes \( R_b + R_r \). The new power dissipated by the bulb is:
\[
P_{\text{new}} = \frac{V^2}{R_b + R_r}
\]

Given that the new power is half of the original power:
\[
P_{\text{new}} = \frac{1}{2} P_{\text{original}}
\]

Substituting the expressions for power:
\[
\frac{V^2}{R_b + R_r} = \frac{1}{2} \cdot \frac{V^2}{R_b}
\]

Simplifying:
\[
\frac{1}{R_b + R_r} = \frac{1}{2R_b}
\]

Cross-multiplying:
\[
2R_b = R_b + R_r
\]

Solving for \( R_r \):
\[
R_r = R_b
\]

#### 1. Potential difference (voltage) across the bulb:
The voltage across the bulb remains the same as the original voltage \( V \) because the total voltage is divided between the bulb and the resistor in series. Therefore:
\[
V_{\text{bulb}} = V
\]

#### 2. Resistance established in the circuit:
The resistance added in series is equal to the resistance of the bulb:
\[
R_r = R_b
\]

#### 3. Power lost in the resistor:
The total current in the circuit is:
\[
I = \frac{V}{R_b + R_r} = \frac{V}{R_b + R_b} = \frac{V}{2R_b}
\]

The power dissipated in the resistor is:
\[
P_{\text{resistor}} = I^2 R_r = \left( \frac{V}{2R_b} \right)^2 R_b = \frac{V^2}{4R_b} \cdot R_b = \frac{V^2}{4R_b}
\]

Since the original power of the bulb was:
\[
P_{\text{original}} = \frac{V^2}{R_b}
\]

The power lost in the resistor is:
\[
P_{\text{resistor}} = \frac{1}{2} \cdot \frac{1}{2} P_{\text{original}} = \frac{1}{4} P_{\text{original}}
\]

#### Final Answers:
1. Potential difference across the bulb: \(\boxed{V}\)
2. Resistance established in the circuit: \(\boxed{R_b}\)
3. Power lost in the resistor: \(\boxed{\frac{1}{4} P_{\text{original}}}\)

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Question 13:

Statement: A certain apparatus operates on a 220 V under which it absorbs 120 W. Under what voltage will it absorb 180 W?

#### Solution:
The power absorbed by the apparatus is given by:
\[
P = \frac{V^2}{R}
\]
where \( R \) is the resistance of the apparatus.

For the first case:
\[
P_1 = 120 \, \text{W}, \quad V_1 = 220 \, \text{V}
\]
\[
R = \frac{V_1^2}{P_1} = \frac{(220 \, \text{V})^2}{120 \, \text{W}} = \frac{48400 \, \text{V}^2}{120 \, \text{W}} = 403.33 \, \Omega
\]

For the second case, let the new voltage be \( V_2 \) and the power be \( P_2 = 180 \, \text{W} \):
\[
P_2 = \frac{V_2^2}{R}
\]
Substituting \( R = 403.33 \, \Omega \):
\[
180 \, \text{W} = \frac{V_2^2}{403.33 \, \Omega}
\]

Solving for \( V_2 \):
\[
V_2^2 = 180 \, \text{W} \times 403.33 \, \Omega = 72600 \, \text{V}^2
\]
\[
V_2 = \sqrt{72600} \approx 269.4 \, \text{V}
\]

#### Final Answer:
\[
\boxed{269.4 \, \text{V}}
\]

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Question 14:

Statement: In a circuit consisting of three elements connected in series, a voltage of 220 V is applied across them. If the voltage drop across element A is 120 V, what is the current through these three elements?

#### Solution:
In a series circuit, the total voltage is the sum of the voltage drops across each element. Let the voltage drops across elements B and C be \( V_B \) and \( V_C \), respectively. The total voltage is:
\[
V_{\text{total}} = V_A + V_B + V_C
\]
Given:
- Total voltage, \( V_{\text{total}} = 220 \, \text{V} \),
- Voltage drop across element A, \( V_A = 120 \, \text{V} \).

Thus:
\[
220 \, \text{V} = 120 \, \text{V} + V_B + V_C
\]
\[
V_B + V_C = 100 \, \text{V}
\]

In a series circuit, the current is the same through all elements. Let the current be \( I \). Using Ohm's Law for element A:
\[
I = \frac{V_A}{R_A}
\]
However, since the resistances of elements B and C are not given, we cannot determine the individual currents or resistances. But the current through the entire circuit is the same as the current through element A.

If additional information about the resistances were provided, we could calculate the exact current. Without it, we can only state that the current is consistent throughout the series circuit.

#### Final Answer:
\[
\boxed{\text{Cannot be determined without additional information}}
\]

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Final Summary of Answers:

1. Question 9: \(\boxed{4 \, \Omega}\)
2. Question 10:
- Resistance: \(\boxed{40 \, \Omega}\)
- Power: \(\boxed{3.6 \, \text{W}}\)
3. Question 11:
- Resistance: \(\boxed{240 \, \Omega}\)
- Current: \(\boxed{0.5 \, \text{A}}\)
4. Question 12:
- Voltage across bulb: \(\boxed{V}\)
- Resistance added: \(\boxed{R_b}\)
- Power lost in resistor: \(\boxed{\frac{1}{4} P_{\text{original}}}\)
5. Question 13: \(\boxed{269.4 \, \text{V}}\)
6. Question 14: \(\boxed{\text{Cannot be determined without additional information}}\)