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.

Question

Asked on June 3, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-03T03:32:55.191Z

Question 1: 1.1.1: Step 1: Calculate the voltage across the parallel combination. Since R_2 is in parallel with R_1, the voltage across R_2 is the same as the voltage across the parallel combination. V_p = I_2 × R_2 = 1 \, A × 6 \, = 6 \, V Step 2: Calculate the current through R_1. I_1 = (V_p)/(R_1) = (6 \, V)/(12 \, ) = 0.5 \, A Step 3: Calculate the total current (I_total) flowing from the supply. 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. I_total = I_1 + I_2 = 0.5 \, A + 1 \, A = 1.5 \, A Step 4: Calculate the total resistance of the circuit. Using Ohm's Law for the entire circuit: R_total = V_supplyI_total = (15 \, V)/(1.5 \, A) = 10 \, Step 5: Calculate the equivalent resistance of the parallel combination (R_p). (1)/(R_p) = (1)/(R_1) + (1)/(R_2) = (1)/(12 \, ) + (1)/(6 \, ) = (1+2)/(12 \, ) = (3)/(12 \, ) = (1)/(4 \, ) R_p = 4 \, Step 6: Calculate the value of the unknown resistor R_x. The total resistance is the sum of the series resistances: R_total = R_p + R_3 + R_x 10 \, = 4 \, + 10 \, + R_x 10 \, = 14 \, + R_x R_x = 10 \, - 14 \, = -4 \, 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 × R_3 = 1.5 \, A × 10 \, = 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 × R_3 = 1.5 \, A × 10 \, = 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 × R_x, then R_x = (V_x)/(I_total) = (-6 \, V)/(1.5 \, A) = -4 \, . 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. Step 1: Calculate the voltage across the parallel combination. V_p = I_2 × R_2 = 1 \, A × 6 \, = 6 \, V Step 2: Calculate the current through R_1. I_1 = (V_p)/(R_1) = (6 \, V)/(12 \, ) = 0.5 \, A Step 3: Calculate the total current (I_total) flowing from the supply. I_total = I_1 + I_2 = 0.5 \, A + 1 \, A = 1.5 \, A Step 4: Calculate the voltage across R_3. V_3 = I_total × R_3 = 1.5 \, A × 10 \, = 15 \, V Step 5: Calculate the voltage across R_x. For a series circuit, the sum of voltages across components equals the supply voltage. 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 Step 6: Calculate the value of R_x. R_x = (V_x)/(I_total) = (-6 \, V)/(1.5 \, A) = -4 \, 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: Step 1: Determine the new total resistance when R_3 is short-circuited. When R_3 is short-circuited, its resistance becomes 0 \, . The equivalent resistance of the parallel combination (R_p) remains 4 \, (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): R'_total = 4 \, + (-4 \, ) = 0 \, If R_x were a positive value, say R_x = 2 \, (hypothetically, to make the original problem consistent, e.g., if V_supply was 23 \, V), then R'_total = 4 \, + 2 \, = 6 \, . 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 \, , then V_x = 1.5 \, A × 2 \, = 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. Step 2: Explain the effect on total current. 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. Step 3: Calculate the new total current. 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 \, ) from the series path will reduce the total resistance. Let's assume for the sake of calculation that R_x was 0 \, 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 \, 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 \, is removed from the series path, R'_total = R_total - 10 \, . Therefore, the total resistance of the circuit will decrease. Step 4: Conclude the effect on total current. According to Ohm's Law, I = (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 \, resistor (R_3) is short-circuited, its resistance effectively becomes 0 \, . 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. → C. ties 2.2.2: An economical method to hold up lines having a span of less than 200 m. → G. wooden poles 2.2.3: A pole that is placed at an angle to hold the main pole in position. → F. strut 2.2.4: Provide a path for surges to be discharged to earth before flashover on the line takes place. → B. lightning arresters 2.2.5: Used to protect, control and monitor LV distribution systems. → 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. Drop the next question. 📸

Accepted Answer · 2026-06-03T03:32:55.191Z

Question 1: 1.1.1: Step 1: Calculate the voltage across the parallel combination. Since R_2 is in parallel with R_1, the voltage across R_2 is the same as the voltage across the parallel combination. V_p = I_2 × R_2 = 1 \, A × 6 \, = 6 \, V Step 2: Calculate the current through R_1. I_1 = (V_p)/(R_1) = (6 \, V)/(12 \, ) = 0.5 \, A Step 3: Calculate the total current (I_total) flowing from the supply. 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. I_total = I_1 + I_2 = 0.5 \, A + 1 \, A = 1.5 \, A Step 4: Calculate the total resistance of the circuit. Using Ohm's Law for the entire circuit: R_total = V_supplyI_total = (15 \, V)/(1.5 \, A) = 10 \, Step 5: Calculate the equivalent resistance of the parallel combination (R_p). (1)/(R_p) = (1)/(R_1) + (1)/(R_2) = (1)/(12 \, ) + (1)/(6 \, ) = (1+2)/(12 \, ) = (3)/(12 \, ) = (1)/(4 \, ) R_p = 4 \, Step 6: Calculate the value of the unknown resistor R_x. The total resistance is the sum of the series resistances: R_total = R_p + R_3 + R_x 10 \, = 4 \, + 10 \, + R_x 10 \, = 14 \, + R_x R_x = 10 \, - 14 \, = -4 \, 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 × R_3 = 1.5 \, A × 10 \, = 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 × R_3 = 1.5 \, A × 10 \, = 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 × R_x, then R_x = (V_x)/(I_total) = (-6 \, V)/(1.5 \, A) = -4 \, . 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. Step 1: Calculate the voltage across the parallel combination. V_p = I_2 × R_2 = 1 \, A × 6 \, = 6 \, V Step 2: Calculate the current through R_1. I_1 = (V_p)/(R_1) = (6 \, V)/(12 \, ) = 0.5 \, A Step 3: Calculate the total current (I_total) flowing from the supply. I_total = I_1 + I_2 = 0.5 \, A + 1 \, A = 1.5 \, A Step 4: Calculate the voltage across R_3. V_3 = I_total × R_3 = 1.5 \, A × 10 \, = 15 \, V Step 5: Calculate the voltage across R_x. For a series circuit, the sum of voltages across components equals the supply voltage. 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 Step 6: Calculate the value of R_x. R_x = (V_x)/(I_total) = (-6 \, V)/(1.5 \, A) = -4 \, 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: Step 1: Determine the new total resistance when R_3 is short-circuited. When R_3 is short-circuited, its resistance becomes 0 \, . The equivalent resistance of the parallel combination (R_p) remains 4 \, (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): R'_total = 4 \, + (-4 \, ) = 0 \, If R_x were a positive value, say R_x = 2 \, (hypothetically, to make the original problem consistent, e.g., if V_supply was 23 \, V), then R'_total = 4 \, + 2 \, = 6 \, . 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 \, , then V_x = 1.5 \, A × 2 \, = 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. Step 2: Explain the effect on total current. 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. Step 3: Calculate the new total current. 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 \, ) from the series path will reduce the total resistance. Let's assume for the sake of calculation that R_x was 0 \, 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 \, 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 \, is removed from the series path, R'_total = R_total - 10 \, . Therefore, the total resistance of the circuit will decrease. Step 4: Conclude the effect on total current. According to Ohm's Law, I = (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 \, resistor (R_3) is short-circuited, its resistance effectively becomes 0 \, . 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. → C. ties 2.2.2: An economical method to hold up lines having a span of less than 200 m. → G. wooden poles 2.2.3: A pole that is placed at an angle to hold the main pole in position. → F. strut 2.2.4: Provide a path for surges to be discharged to earth before flashover on the line takes place. → B. lightning arresters 2.2.5: Used to protect, control and monitor LV distribution systems. → 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. Drop the next question. 📸

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.

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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.

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