Ohm's Law Worksheet with 10 practice problems on voltage, current, and resistance.

It looks like you've uploaded a worksheet titled "Ohm's Law Worksheet" with several problems related to Ohm's Law, which is expressed by the formula:

\[
V = I \times R
\]

Where:
- \( V \) = Voltage (in volts, V)
- \( I \) = Current (in amperes, A)
- \( R \) = Resistance (in ohms, Ω)

Let’s go through each question step by step and solve them.

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Question 1:

How much current flows through a circuit that includes a 9-volt battery and a bulb with a resistance of 18 ohms?

Use Ohm’s Law:
\[
I = \frac{V}{R} = \frac{9\,V}{18\,\Omega} = 0.5\,A
\]

✔ Answer: 0.5 A

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Question 2:

A 12-volt car battery is connected to a 30-ohm resistor. What is the current flowing in the circuit?

\[
I = \frac{V}{R} = \frac{12\,V}{30\,\Omega} = 0.4\,A
\]

✔ Answer: 0.4 A

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Question 3:

A 2.2-kilohm resistor has 18 mA current passing through it. Find the value of the voltage across the resistor.

First, convert units:
- 2.2 kΩ = 2200 Ω
- 18 mA = 0.018 A

Now use:
\[
V = I \times R = 0.018\,A \times 2200\,\Omega = 39.6\,V
\]

✔ Answer: 39.6 V

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Question 4:

A small candle (2-watt battery) is connected to a light bulb having resistance of 2 ohms. Find the current.

Wait — "2-watt battery"? That might be a typo or misstatement. Batteries are usually rated in volts, not watts. But if it's a 2-watt device, we can still work with power.

But here, it says “a small candle (2-watt battery)” — likely meant to say “a 2-watt bulb” or “a 2-volt battery”.

Assuming it's a 2-volt battery, then:

\[
I = \frac{V}{R} = \frac{2\,V}{2\,\Omega} = 1\,A
\]

Alternatively, if it's a 2-watt bulb with 2 ohms resistance, then:

Use Power formula:
\[
P = I^2 R \Rightarrow I = \sqrt{\frac{P}{R}} = \sqrt{\frac{2}{2}} = 1\,A
\]

Same result.

✔ Answer: 1 A

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Question 5:

Two resistors, one of 3.7 ohm and other one of 12.7 ohm, are connected in series to a 12-volt battery. Find the current through the circuit.

In series, total resistance:
\[
R_{total} = 3.7 + 12.7 = 16.4\,\Omega
\]

Current:
\[
I = \frac{V}{R} = \frac{12\,V}{16.4\,\Omega} \approx 0.732\,A
\]

✔ Answer: ≈ 0.732 A

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Question 6:

Two resistors, one of 10 ohm and other of 15 ohm, are connected in parallel to a 12-volt battery. What is the current through each resistor?

In parallel, voltage across each resistor is 12 V.

For 10 Ω:
\[
I_1 = \frac{12\,V}{10\,\Omega} = 1.2\,A
\]

For 15 Ω:
\[
I_2 = \frac{12\,V}{15\,\Omega} = 0.8\,A
\]

✔ Answer: 1.2 A through 10 Ω, 0.8 A through 15 Ω

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Question 7:

How current changes in circuit for constant voltage, when resistance value increases or decreases?

From Ohm’s Law: \( I = \frac{V}{R} \)

- If resistance increases, current decreases (inverse relationship).
- If resistance decreases, current increases.

✔ Answer: Current decreases when resistance increases; current increases when resistance decreases.

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Question 8:

Calculate how long it takes to charge a capacitor of 200 μF with a 5-V supply through a 100 kΩ resistor.

This is a RC charging time problem.

Time constant:
\[
\tau = R \times C = 100,000\,\Omega \times 200 \times 10^{-6}\,F = 20\,seconds
\]

The capacitor charges to about 63% of full voltage in one time constant. To reach ~99%, it takes about 5τ.

So:
\[
t = 5 \times \tau = 5 \times 20 = 100\,seconds
\]

✔ Answer: Approximately 100 seconds

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Question 9:

A 5-V DC source is connected to a 10 Ω resistor. What is the power dissipated?

Use:
\[
P = \frac{V^2}{R} = \frac{(5)^2}{10} = \frac{25}{10} = 2.5\,W
\]

Or:
\[
I = \frac{5}{10} = 0.5\,A \Rightarrow P = I \times V = 0.5 \times 5 = 2.5\,W
\]

✔ Answer: 2.5 W

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Question 10:

Find the value of resistance \( R \) such that the current in the circuit is 2.5 A when a 10-V source is applied.

\[
R = \frac{V}{I} = \frac{10\,V}{2.5\,A} = 4\,\Omega
\]

✔ Answer: 4 Ω

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Question 11:

A circuit has a 12-V battery and a 4-ohm resistor. Calculate the power dissipated in the resistor.

\[
I = \frac{12}{4} = 3\,A
\]
\[
P = I^2 R = (3)^2 \times 4 = 9 \times 4 = 36\,W
\]

Or:
\[
P = \frac{V^2}{R} = \frac{144}{4} = 36\,W
\]

✔ Answer: 36 W

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Final Summary of Answers:

| Question | Answer |
|--------|--------|
| 1 | 0.5 A |
| 2 | 0.4 A |
| 3 | 39.6 V |
| 4 | 1 A |
| 5 | ~0.732 A |
| 6 | 1.2 A and 0.8 A |
| 7 | Current decreases with increasing resistance; increases with decreasing resistance |
| 8 | ~100 seconds |
| 9 | 2.5 W |
| 10 | 4 Ω |
| 11 | 36 W |

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