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### a) What is kilowatt-hour?

**Kilowatt-hour (kWh)** is a unit of energy. It represents the amount of energy consumed when a device with a power of one kilowatt operates for one hour. It is commonly used as a billing unit for energy delivered to consumers by electric utilities.

1 kilowatt-hour = 1000 watts * 1 hour = 3,600,000 joules.

### b) What would the current versus potential difference graph look like in Ohm's law? Explain.

In Ohm's law, the current (I) through a conductor between two points is directly proportional to the potential difference (V) across the two points. The relationship is given by:

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

Where \( R \) is the resistance. If we plot a graph of current (I) on the y-axis versus potential difference (V) on the x-axis, it would be a straight line passing through the origin with a slope equal to \( \frac{1}{R} \). The slope of the line depends on the resistance. If the resistance is constant, the line will be straight.

### c) Determine the equivalent resistance of the circuit.

The circuit consists of three resistors:

- \( R_1 = 100 \ \Omega \)
- \( R_2 = 100 \ \Omega \)
- \( R_3 = 200 \ \Omega \)
- \( R_4 = 100 \ \Omega \)

First, let's identify the configuration of these resistors. The circuit looks like a combination of series and parallel resistances:

1. \( R_1 \) and \( R_2 \) are in series. Their combined resistance:
   \[
   R_{12} = R_1 + R_2 = 100 \ \Omega + 100 \ \Omega = 200 \ \Omega
   \]

2. \( R_3 \) and \( R_4 \) are also in series. Their combined resistance:
   \[
   R_{34} = R_3 + R_4 = 200 \ \Omega + 100 \ \Omega = 300 \ \Omega
   \]

3. Now, \( R_{12} \) and \( R_{34} \) are in parallel. The equivalent resistance \( R_{eq} \) is calculated using:
   \[
   \frac{1}{R_{eq}} = \frac{1}{R_{12}} + \frac{1}{R_{34}} = \frac{1}{200 \ \Omega} + \frac{1}{300 \ \Omega}
   \]
   \[
   \frac{1}{R_{eq}} = \frac{3 + 2}{600} = \frac{5}{600} = \frac{1}{120} \ \Omega
   \]
   So,
   \[
   R_{eq} = 120 \ \Omega
   \]

### d) Analyze mathematically the change in current flow when arm AB is removed (including resistance R₄).

If arm AB (which likely includes \( R_4 \)) is removed, we must reconsider the configuration of the circuit:

- Without \( R_4 \), only \( R_1 \), \( R_2 \), and \( R_3 \) remain.
- \( R_1 \) and \( R_2 \) are still in series, so \( R_{12} = 200 \ \Omega \).
- Since \( R_3 \) is now alone, the circuit reduces to \( R_{12} \) in parallel with \( R_3 \).

The new equivalent resistance will be:
\[
\frac{1}{R'_{eq}} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{200 \ \Omega} + \frac{1}{200 \ \Omega} = \frac{2}{200} = \frac{1}{100} \ \Omega
\]
\[
R'_{eq} = 100 \ \Omega
\]

When \( R_4 \) is removed, the equivalent resistance decreases from 120 Ω to 100 Ω, meaning the overall current in the circuit will increase due to the decrease in resistance.

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Would you like more details or have any questions? Here are some related questions:

1. How is electrical energy consumption calculated using kilowatt-hours?
2. What happens to the slope of the I-V graph if the resistance is not constant?
3. How can the equivalent resistance be calculated for more complex circuits?
4. How does removing a resistor from a parallel circuit affect the total current?
5. What would happen to the circuit's power dissipation if \( R_4 \) were changed instead of removed?

**Tip:** Always double-check the circuit configuration before calculating the equivalent resistance to avoid mistakes!

a) What is kilowatt-hour?
b) What would the current versus potential difference graph look like in Ohm's law? Explain.
c) Determine the equivalent resistance of the circuit.
d) Analyze mathematically the change in current flow when arm AB is removed (including resistance R₄).

Explore the concept of kilowatt-hour, how current versus potential difference graphs look under Ohm's law, and how to calculate equivalent resistance in series-parallel circuits. Also, learn the impact on current flow when a resistor is removed.

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