Four electrical circuit problems involving voltage, resistance, and current calculations.

Let's solve each question step by step.

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Question 4:

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

#### Solution:
We use Ohm's Law, which states:
\[
I = \frac{V}{R}
\]
where:
- \( I \) is the current (in amperes),
- \( V \) is the voltage (in volts),
- \( R \) is the resistance (in ohms).

Given:
- \( V = 12 \) V,
- \( R = 5 \) Ω.

Substitute the values into Ohm's Law:
\[
I = \frac{12}{5} = 2.4 \text{ A}
\]

#### Answer:
\[
\boxed{2.4 \text{ A}}
\]

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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 kohm. 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 kohm, which is:
\[
R = 1 \text{ kΩ} = 1000 \text{ Ω}
\]

Using Ohm's Law:
\[
I = \frac{V_{\text{total}}}{R} = \frac{15}{1000} = 0.015 \text{ A}
\]

#### Answer:
\[
\boxed{0.015 \text{ A}}
\]

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Question 6:

Two lamps, each having a resistance of 3 ohms, are connected 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:
\[
R_{\text{total}} = 3 \text{ Ω} + 3 \text{ Ω} = 6 \text{ Ω}
\]

The voltage \( V \) is given as 5 V. Using Ohm's Law:
\[
I = \frac{V}{R_{\text{total}}} = \frac{5}{6} \approx 0.833 \text{ A}
\]

#### Answer:
\[
\boxed{0.833 \text{ A}}
\]

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Question 7:

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

#### Solution:
Ohm's Law states:
\[
I = \frac{V}{R}
\]
where:
- \( I \) is the current,
- \( V \) is the voltage (constant in this case),
- \( R \) is the resistance.

If the voltage \( V \) is constant and the resistance \( R \) increases, the current \( I \) decreases because \( I \) is inversely proportional to \( R \). Mathematically:
\[
I \propto \frac{1}{R}
\]

#### Answer:
\[
\boxed{\text{Current decreases}}
\]

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Question 8:

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

#### Solution:
We use Ohm's Law:
\[
R = \frac{V}{I}
\]
where:
- \( V = 50 \) V,
- \( I = 10 \) A.

Substitute the values:
\[
R = \frac{50}{10} = 5 \text{ Ω}
\]

#### Answer:
\[
\boxed{5 \text{ Ω}}
\]

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Final Answers:

1. Question 4: \(\boxed{2.4 \text{ A}}\)
2. Question 5: \(\boxed{0.015 \text{ A}}\)
3. Question 6: \(\boxed{0.833 \text{ A}}\)
4. Question 7: \(\boxed{\text{Current decreases}}\)
5. Question 8: \(\boxed{5 \text{ Ω}}\)