Ohm’s law states that the current flowing in a fixed resistor is directly proportional to the applied voltage. When 30 volts is applied across a resistor the current flowing through the resistor is 2.4×10−3 amperes. Determine

 

(a) the constant of proportionality,

(b) the current when the voltage is 52 volts and

(c) the voltage when the current is 3.6×10−3 amperes.

Ohm's Law can be expressed as:

\[
V = IR
\]
where:
- \( V \) is the voltage (in volts),
- \( I \) is the current (in amperes),
- \( R \) is the resistance (in ohms).

We can rearrange this equation to solve for the resistance \( R \):

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

Now, let's go through the problem step by step.

### (a) The constant of proportionality (resistance \( R \)):

Given:
- Voltage \( V = 30 \, \text{V} \),
- Current \( I = 2.4 \times 10^{-3} \, \text{A} \).

Using the formula \( R = \frac{V}{I} \):

\[
R = \frac{30}{2.4 \times 10^{-3}} = 12500 \, \Omega
\]

So, the constant of proportionality (resistance \( R \)) is \( 12500 \, \Omega \).

---

### (b) The current when the voltage is 52 volts:

We already know the resistance \( R = 12500 \, \Omega \), and the voltage \( V = 52 \, \text{V} \).

Using the formula \( I = \frac{V}{R} \):

\[
I = \frac{52}{12500} = 4.16 \times 10^{-3} \, \text{A}
\]

So, the current when the voltage is 52 volts is \( 4.16 \times 10^{-3} \, \text{A} \).

---

### (c) The voltage when the current is \( 3.6 \times 10^{-3} \, \text{A} \):

Given:
- Current \( I = 3.6 \times 10^{-3} \, \text{A} \),
- Resistance \( R = 12500 \, \Omega \).

Using the formula \( V = IR \):

\[
V = 3.6 \times 10^{-3} \times 12500 = 45 \, \text{V}
\]

So, the voltage when the current is \( 3.6 \times 10^{-3} \, \text{A} \) is \( 45 \, \text{V} \).

---

### Summary:

- (a) The constant of proportionality \( R = 12500 \, \Omega \),
- (b) The current when the voltage is 52 volts is \( 4.16 \times 10^{-3} \, \text{A} \),
- (c) The voltage when the current is \( 3.6 \times 10^{-3} \, \text{A} \) is \( 45 \, \text{V} \).

Would you like further clarification or additional details?

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Here are 5 related questions to expand your understanding:
1. How does the resistance affect the current in a circuit if the voltage is kept constant?
2. What is the power dissipated by the resistor in part (a)?
3. How would the results change if the resistor’s resistance increased?
4. What would be the current if the voltage were doubled from 30 volts?
5. Can you explain how Ohm's law relates to energy consumption in electrical devices?

**Tip**: Always ensure units are consistent when applying formulas in electrical problems (e.g., amperes for current, volts for voltage, ohms for resistance).

Ohm’s law states that the current flowing in a fixed resistor is directly proportional to the applied voltage. When 30 volts is applied across a resistor the current flowing through the resistor is 2.4×10−3 amperes. Determine (a) the constant of proportionality, (b) the current when the voltage is 52 volts and (c) the voltage when the current is 3.6×10−3 amperes.

Learn how to apply Ohm’s law to solve problems involving resistance, current, and voltage in electrical circuits. This problem covers key concepts like Ohm's Law, direct proportionality, and step-by-step calculations of resistance (12500 ohms), current (4.16×10−3 amperes), and voltage (45 volts). Suitable for students in Grades 9-12.

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