Question

Power source that provides a current of 1.6 A to two 5 Ω resistors connected in series is moved to a parallel circuit that consists of three identical resistors. In the parallel circuit, the overall current is 2.0 A.
The value of a resistor used in the parallel circuit is?

A)5.3Ω
B)8.0Ω
C)12Ω
D)24Ω

It's D :P

Answer 1

This one is complicated. Not hard, but several steps. The two 5-ohm resistors in series have a total resistance of 10 ohms. 1.6 A through 10 ohms means the battery is (1.6 x 10)=16 volts. Then ... 16 volts produces 2A of current through (16/2)=8 ohms of resistance. If 3 identical resistors in parallel look like like 8 ohms, then each one is (8 x 3)= 24 ohms. That's D.

Answer 2

Answer: The correct answer is Option D.

Explanation:

To calculate the value of resistor used in parallel circuit, we need to follow some steps:

• Step 1: Calculation of Voltage from series circuit.

For the calculation of voltage, we will use thew equation given by Ohm's Law, which is:

$V=IR$      ....(1)

where,

V = voltage = ?V

I = Current flowing = 1.6 A

R = Resistance = (5 + 5) = 10 ohm   (In Series, resistance gets added up)

Putting values in above equation, we get:

$V=1.6\times 10=16V$

• Step 2: Finding the total resistance from parallel circuit

As, the voltage passing through the resistors remains the same. Hence, the above voltage can be used:

I = 2A

V = 16V

Putting values in equation 1, we get:

$16=2\times R_{parallel}\\\\R_{parallel}=8\Omega$

• Step 3: Finding the value of each resistor.

We are given 3 identical resistors that are connected in parallel. In parallel, the total resistance is calculated by using the formula:

$\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\\R_1=R_2=R_3=R\\\\\frac{1}{R_{parallel}}=\frac{1+1+1}{R}=\frac{3}{R}$

Putting values in above equation, we get:

$\frac{1}{8}=\frac{3}{R}\\\\R=24\Omega$

Hence, the correct answer is Option D.