Question

Two 24Ω resistors are in parallel, and a 42Ω resistor is in series with the combination. When 78V is applied to the three resistors, the voltage drop across the 42Ω resistor is about____

a. 49.8 V
b. 65.3 V
c. 55.8 V
d. 60.5 V

Answer 1

The two resistors in parallel will have a combined reistance of 12 Ω.

The total resistance of the circuit is then

R = (12 + 42) Ω = 54 Ω

I = V / R = 78 V / 54 Ω = 1.44 amps

The voltage drop across 42 Ω is then

V42 = 42 Ω * 1.44 amps = 60.5 V

(D) is correct

Answer 2

By calculating the equivalent resistance of two 24Ω resistors in parallel and adding it to the series 42Ω resistor, the total current in the circuit is found with Ohm's Law, and thus the voltage drop across the 42Ω resistor is approximately 60.5V.

Step-by-step explanation:

To find the voltage drop across the 42Ω resistor, we first need to calculate the equivalent resistance of the two 24Ω resistors in parallel. The equivalent resistance (Req) of two resistors in parallel can be calculated using the formula:

1/Req = 1/R1 + 1/R2

For two identical 24Ω resistors in parallel, this becomes:

1/Req = 1/24 + 1/24 = 2/24
Req = 24/2 = 12Ω

Now, the total resistance of the circuit, which is the sum of the equivalent resistance of the parallel combination and the series 42Ω resistor, is:

Rtotal = Req + R3
Rtotal = 12Ω + 42Ω = 54Ω

Next, using Ohm's law (V = I·R), we can find the total current (Itotal) flowing through the circuit as follows:

Itotal = Vsource / Rtotal
Itotal = 78V / 54Ω = 1.44 A

Lastly, to find the voltage drop across the 42Ω resistor:

VR3 = Itotal · R3
VR3 = 1.44 A · 42Ω ≈ 60.5V

Therefore, the voltage drop across the 42Ω resistor when a 78V voltage is applied to the circuit is about 60.5V.