QUESTION 2: Discuss the effect that voltage has on transmission efficiency. Choose a term from COLUMN B that matches a description in COLUMN A. Write only the letter (A–H) next to the question number (2.2.1–2.2.5) in the ANSWER BOOK.

Since $R_2$ is in parallel with $R_1$, the voltage across $R_2$ is the same as the voltage across the parallel combination.

The total current is the sum of currents in the parallel branches. This current also flows through the series resistors $R_3$ and $R_x$.

Using Ohm's Law for the entire circuit:

The total resistance is the sum of the series resistances:

There seems to be an issue with the problem statement or the given values, as a resistance cannot be negative. Let's re-check the calculation.

Ah, I see the error in my calculation for $R_{total}$. The total resistance is $R_p + R_3 + R_x$.
The voltage across the parallel combination is $V_p = 6 \, V$.
The voltage across $R_3$ is $V_3 = I_{total} \times R_3 = 1.5 \, A \times 10 \, \Omega = 15 \, V$.
The total supply voltage is $15 \, V$.
So, $V_{supply} = V_p + V_3 + V_x$.
$15 \, V = 6 \, V + 15 \, V + V_x$.
$15 \, V = 21 \, V + V_x$.
$V_x = 15 \, V - 21 \, V = -6 \, V$.
This still leads to a negative voltage across $R_x$, implying a negative resistance. This indicates that the given values in the problem are inconsistent. Specifically, the sum of voltages across the parallel part and $R_3$ already exceeds the supply voltage.

Let's assume the question implies that $R_3$ and $R_x$ are in series with the parallel combination, and the total voltage across this series combination is $15 \, V$.
The voltage across the parallel part is $V_p = 6 \, V$.
The voltage across $R_3$ is $V_3 = I_{total} \times R_3 = 1.5 \, A \times 10 \, \Omega = 15 \, V$.
The sum of these voltages is $V_p + V_3 = 6 \, V + 15 \, V = 21 \, V$.
This sum ($21 \, V$) is already greater than the total supply voltage ($15 \, V$). This means the circuit as described is not physically possible with the given values.

Given the inconsistency, I will proceed by assuming there might be a typo in the supply voltage or the current through $R_2$. However, I must use the given values. If $V_{supply} = V_p + V_3 + V_x$, then $V_x = V_{supply} - V_p - V_3 = 15 \, V - 6 \, V - 15 \, V = -6 \, V$.
Since $V_x = I_{total} \times R_x$, then $R_x = \frac{V_x}{I_{total}} = \frac{-6 \, V}{1.5 \, A} = -4 \, \Omega$.

As resistance cannot be negative, there is an error in the problem statement's values. I will state this and provide the calculation based on the given (inconsistent) values.

For a series circuit, the sum of voltages across components equals the supply voltage.

Note: The calculated resistance $R_x$ is negative, which is not physically possible. This indicates an inconsistency in the problem's given values.

1.1.2:

When $R_3$ is short-circuited, its resistance becomes $0 \, \Omega$.
The equivalent resistance of the parallel combination ($R_p$) remains $4 \, \Omega$ (calculated in 1.1.1).
The new total resistance $R'_{total} = R_p + R_x$.
Using the value of $R_x$ from 1.1.1 (even though it's negative due to inconsistency):

If $R_x$ were a positive value, say $R_x = 2 \, \Omega$ (hypothetically, to make the original problem consistent, e.g., if $V_{supply}$ was $23 \, V$), then $R'_{total} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega$.
Let's assume the question expects us to consider $R_x$ as a positive value if the problem were consistent. If $R_x$ was intended to be a positive value, then short-circuiting $R_3$ would decrease the total resistance.

Let's re-evaluate 1.1.1 assuming $R_x$ is positive and the supply voltage is adjusted to make the problem consistent. If $R_x$ was, for example, $2 \, \Omega$, then $V_x = 1.5 \, A \times 2 \, \Omega = 3 \, V$. Then $V_{supply} = 6 \, V + 15 \, V + 3 \, V = 24 \, V$.
However, I must use the given values. The problem is fundamentally flawed with the given numbers.

Let's assume the question intends for $R_x$ to be a positive value, and the inconsistency is a mistake in the problem statement. If $R_3$ is short-circuited, the total resistance of the circuit decreases because one of the series resistors is removed.
Original total resistance (if $R_x$ was positive): $R_{total} = R_p + R_3 + R_x$.
New total resistance: $R'_{total} = R_p + R_x$.
Since $R_3$ is removed, $R'_{total} < R_{total}$.

According to Ohm's Law ($I = V/R$), if the total resistance ($R'_{total}$) decreases while the supply voltage ($V_{supply}$) remains constant, the total current will increase.

Since the problem values are inconsistent, a precise numerical calculation for $R_x$ is problematic. However, the qualitative effect is clear.
If we assume $R_x$ was intended to be a positive value, then removing $R_3$ (which is $10 \, \Omega$) from the series path will reduce the total resistance.
Let's assume for the sake of calculation that $R_x$ was $0 \, \Omega$ in the original problem, which would make $V_{supply} = V_p + V_3 = 6 \, V + 15 \, V = 21 \, V$. This is still not $15 \, V$.

Given the inconsistency, I will explain the effect qualitatively and then perform a calculation using the original $R_p$ and assuming $R_x$ is some positive value (e.g., $0 \, \Omega$ for simplicity, or just state the general effect).

Let's assume the question implies that the total resistance of the circuit will decrease.
Original total resistance (from 1.1.1, if $R_x$ was positive): $R_{total} = R_p + R_3 + R_x$.
New total resistance (with $R_3$ short-circuited): $R'_{total} = R_p + R_x$.
Since $R_3 = 10 \, \Omega$ is removed from the series path, $R'_{total} = R_{total} - 10 \, \Omega$.
Therefore, the total resistance of the circuit will decrease.

According to Ohm's Law, $I = \frac{V}{R}$. If the supply voltage $V$ remains constant and the total resistance $R$ decreases, the total current $I$ will increase.

• Effect: The total current will increase.
• Substantiation: When the $10 \, \Omega$ resistor ($R_3$) is short-circuited, its resistance effectively becomes $0 \, \Omega$. This reduces the total equivalent resistance of the circuit. Since the supply voltage remains constant, according to Ohm's Law ($I = V/R$), a decrease in total resistance leads to an increase in total current.

Question 2:

2.1:
High voltage significantly improves transmission efficiency. When power is transmitted at higher voltages, the current required to transmit the same amount of power ($P = VI$) is lower. Since power loss in transmission lines is proportional to the square of the current ($P_{loss} = I^2 R$), a lower current drastically reduces energy losses due to the resistance of the transmission cables.

2.2:

• 2.2.1: Used to attach conductors to insulators on overhead systems. $\rightarrow$ C. ties
• 2.2.2: An economical method to hold up lines having a span of less than 200 m. $\rightarrow$ G. wooden poles
• 2.2.3: A pole that is placed at an angle to hold the main pole in position. $\rightarrow$ F. strut
• 2.2.4: Provide a path for surges to be discharged to earth before flashover on the line takes place. $\rightarrow$ B. lightning arresters
• 2.2.5: Used to protect, control and monitor LV distribution systems. $\rightarrow$ D. switchgear

2.3:
A fuse is a safety device designed to protect electrical circuits from overcurrent. It contains a metal wire or filament that melts and breaks the circuit when the current flowing through it exceeds a safe level, preventing damage to appliances or wiring and reducing the risk of fire.