BMST - Basic Math and Science Electrical Circuits and Ohm's Law Questions and Answers 4

Question 1

A resistor network consists of two resistors in parallel: R1 = 8 Ω and R2 = 12 Ω. This parallel combination is then connected in series with R3 = 5 Ω. If the total supply voltage is 24 V, what is the voltage drop across the parallel combination?

Accepted Answer

9.6 V

Answer

First, find the equivalent resistance of the parallel pair: 1/Rp = 1/8 + 1/12 = 3/24 + 2/24 = 5/24, so Rp = 24/5 = 4.8 Ω. Total resistance = 4.8 + 5 = 9.8 Ω. Total current = 24/9.8 ≈ 2.449 A. Voltage across the parallel combination = 2.449 × 4.8 ≈ 9.6 V.

Question 2

A wire carries a current of 3 A through a resistor. The power dissipated is 27 W. If the current is doubled to 6 A while the resistor value stays the same, what is the new power dissipated?

Accepted Answer

108 W

Answer

Power is given by P = I²R. The original resistance R = P/I² = 27/9 = 3 Ω. With I = 6 A, new power = (6)² × 3 = 36 × 3 = 108 W. Since power is proportional to the square of current, doubling the current quadruples the power: 27 × 4 = 108 W.

Question 3

Two batteries are connected in series: Battery A has an EMF of 9 V with an internal resistance of 1 Ω, and Battery B has an EMF of 6 V with an internal resistance of 0.5 Ω. They are connected to an external load of 8.5 Ω. What is the terminal voltage across the external load?

Accepted Answer

12.75 V

Answer

Total EMF = 9 + 6 = 15 V. Total resistance = 1 + 0.5 + 8.5 = 10 Ω. Current = 15/10 = 1.5 A. Voltage across the external load = I × R_load = 1.5 × 8.5 = 12.75 V. Alternatively, terminal voltage = Total EMF − I × (r1 + r2) = 15 − 1.5 × 1.5 = 15 − 2.25 = 12.75 V.

Question 4

A circuit has three resistors in parallel across a 12 V source: R1 = 6 Ω, R2 = 4 Ω, and R3 = 12 Ω. What is the total current drawn from the source?

Accepted Answer

5.5 A

Answer

In a parallel circuit, the voltage across each resistor is equal to the source voltage (12 V). I1 = 12/6 = 2 A, I2 = 12/4 = 3 A, I3 = 12/12 = 1 A. Total current = 2 + 3 + 1 = 6 A... Wait — recalculating: 2 + 3 + 0.5 = 5.5 A. R3 = 12 Ω → I3 = 12/12 = 1 A. Total = 2 + 3 + 1 = 6 A. Correction: answer is 6.0 A — the branch currents are 2 A, 3 A, and 1 A summing to 6 A.

Question 5

A technician measures the resistance of a conductor and finds it is 4 Ω at 20°C. The temperature coefficient of resistance for this material is 0.004 /°C. What is the resistance at 70°C?

Accepted Answer

4.8 Ω

Answer

Using the formula R_T = R_0 × [1 + α(T − T_0)]: R_70 = 4 × [1 + 0.004 × (70 − 20)] = 4 × [1 + 0.004 × 50] = 4 × [1 + 0.2] = 4 × 1.2 = 4.8 Ω. The resistance increases with temperature for most conductors.

BMST - Basic Math and Science Test Practice Test

BMST - Basic Math and Science Test Practice Test

BMST - Basic Math and Science Electrical Circuits and Ohm's Law Questions and Answers 4