Ohms Law Practice Worksheet With Answers [PDF Download Available ... - Free Printable

Ohm's Law Worksheet Solutions

Ohm's Law is a fundamental principle in electrical circuits, which states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor. Mathematically, it is expressed as:

\[
V = I \cdot R
\]

where:
- \( V \) is the voltage (in volts, V),
- \( I \) is the current (in amperes, A),
- \( R \) is the resistance (in ohms, Ω).

We will solve each question step by step.

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Question 1: State Ohm's law?

Solution:
Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor. Mathematically, it is expressed as:

\[
V = I \cdot R
\]

or equivalently,

\[
I = \frac{V}{R} \quad \text{and} \quad R = \frac{V}{I}
\]

Answer: Ohm's Law: \( V = I \cdot R \)

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Question 2: A 5 V source connects to a 10 ohms resistor. What is the current?

Solution:
Using Ohm's Law:

\[
I = \frac{V}{R}
\]

Given:
- \( V = 5 \, \text{V} \)
- \( R = 10 \, \Omega \)

Substitute the values:

\[
I = \frac{5}{10} = 0.5 \, \text{A}
\]

Answer: \( I = 0.5 \, \text{A} \)

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Question 3: A 2.2 kΩ resistor has 15 mA current passing through it. Find the value of the connected voltage source.

Solution:
Using Ohm's Law:

\[
V = I \cdot R
\]

Given:
- \( I = 15 \, \text{mA} = 0.015 \, \text{A} \) (since \( 1 \, \text{mA} = 0.001 \, \text{A} \))
- \( R = 2.2 \, \text{kΩ} = 2200 \, \Omega \) (since \( 1 \, \text{kΩ} = 1000 \, \Omega \))

Substitute the values:

\[
V = 0.015 \cdot 2200 = 33 \, \text{V}
\]

Answer: \( V = 33 \, \text{V} \)

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Question 4: A circuit contains a 12-volt battery connected to a light bulb having a resistance of 5 ohms. Find the current.

Solution:
Using Ohm's Law:

\[
I = \frac{V}{R}
\]

Given:
- \( V = 12 \, \text{V} \)
- \( R = 5 \, \Omega \)

Substitute the values:

\[
I = \frac{12}{5} = 2.4 \, \text{A}
\]

Answer: \( I = 2.4 \, \text{A} \)

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Question 5: Two batteries, one of 3 V and the other of 12 V, are connected in series to a resistor of 1 kΩ. Find the current that will flow through the resistors.

Solution:
When batteries are connected in series, their voltages add up. Therefore, the total voltage \( V_{\text{total}} \) is:

\[
V_{\text{total}} = 3 \, \text{V} + 12 \, \text{V} = 15 \, \text{V}
\]

The resistance \( R \) is given as \( 1 \, \text{kΩ} = 1000 \, \Omega \).

Using Ohm's Law:

\[
I = \frac{V_{\text{total}}}{R}
\]

Substitute the values:

\[
I = \frac{15}{1000} = 0.015 \, \text{A} = 15 \, \text{mA}
\]

Answer: \( I = 15 \, \text{mA} \)

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Question 6: Two lamps, each having a resistance of 3 Ω, connect in series. What current will flow if a voltage source of 5 V is connected at the input?

Solution:
When resistors are connected in series, their resistances add up. Therefore, the total resistance \( R_{\text{total}} \) is:

\[
R_{\text{total}} = 3 \, \Omega + 3 \, \Omega = 6 \, \Omega
\]

The voltage \( V \) is given as \( 5 \, \text{V} \).

Using Ohm's Law:

\[
I = \frac{V}{R_{\text{total}}}
\]

Substitute the values:

\[
I = \frac{5}{6} \approx 0.833 \, \text{A}
\]

Answer: \( I \approx 0.833 \, \text{A} \)

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Question 7: How does current change in a circuit for constant voltage when the resistance value increases?

Solution:
From Ohm's Law:

\[
I = \frac{V}{R}
\]

If the voltage \( V \) is constant and the resistance \( R \) increases, the current \( I \) decreases. This is because current is inversely proportional to resistance.

Answer: Current decreases when resistance increases for a constant voltage.

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Question 8: A certain resistance has 10 A current through it when a 50 V source is applied. Find the value of resistance.

Solution:
Using Ohm's Law:

\[
R = \frac{V}{I}
\]

Given:
- \( V = 50 \, \text{V} \)
- \( I = 10 \, \text{A} \)

Substitute the values:

\[
R = \frac{50}{10} = 5 \, \Omega
\]

Answer: \( R = 5 \, \Omega \)

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Question 9: A 5 V, 3 mA LED connected to a 12 V source requires a series resistor of how many ohms?

Solution:
The LED has a voltage drop of \( 5 \, \text{V} \) and a current of \( 3 \, \text{mA} = 0.003 \, \text{A} \). The total voltage supplied by the source is \( 12 \, \text{V} \).

The voltage across the series resistor \( V_R \) is:

\[
V_R = V_{\text{source}} - V_{\text{LED}} = 12 \, \text{V} - 5 \, \text{V} = 7 \, \text{V}
\]

The current through the resistor is the same as the current through the LED, which is \( I = 3 \, \text{mA} = 0.003 \, \text{A} \).

Using Ohm's Law:

\[
R = \frac{V_R}{I}
\]

Substitute the values:

\[
R = \frac{7}{0.003} = 2333.33 \, \Omega \approx 2.33 \, \text{kΩ}
\]

Answer: \( R \approx 2.33 \, \text{kΩ} \)

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Question 10: Find the current supplied by a 10 V source to two parallel resistors of 6 Ω each.

Solution:
When resistors are connected in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by:

\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}
\]

Given:
- \( R_1 = 6 \, \Omega \)
- \( R_2 = 6 \, \Omega \)

Substitute the values:

\[
\frac{1}{R_{\text{eq}}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}
\]

Thus,

\[
R_{\text{eq}} = 3 \, \Omega
\]

The voltage \( V \) is given as \( 10 \, \text{V} \).

Using Ohm's Law:

\[
I = \frac{V}{R_{\text{eq}}}
\]

Substitute the values:

\[
I = \frac{10}{3} \approx 3.33 \, \text{A}
\]

Answer: \( I \approx 3.33 \, \text{A} \)

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Final Answers:

1. \( \boxed{V = I \cdot R} \)
2. \( \boxed{0.5 \, \text{A}} \)
3. \( \boxed{33 \, \text{V}} \)
4. \( \boxed{2.4 \, \text{A}} \)
5. \( \boxed{15 \, \text{mA}} \)
6. \( \boxed{0.833 \, \text{A}} \)
7. Current decreases.
8. \( \boxed{5 \, \Omega} \)
9. \( \boxed{2.33 \, \text{kΩ}} \)
10. \( \boxed{3.33 \, \text{A}} \)