Question

A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel across a 24.0 V battery. What is the equivalent resistance of the circuit? Answer in units of Ω.
What is the current in the 4.0 Ω resistor? Answer in units of A.
What is the current in the 8.0 Ω resistor? Answer in units of A.

Answer 1

PART A)

Equivalent resistance in parallel is given as

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

now we have

$\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12}$

$R = 2.18 ohm$

PART B)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

$V = iR$

for 4 ohm resistance we have

$24 = 4(i)$

$i = 6 A$

PART C)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

$V = iR$

for 8 ohm resistance we have

$24 = 8(i)$

$i = 3 A$

Answer 2

Considering the definition of parallel resistor and Ohm's law:

• The equivalent resistance of the circuit is 2.18 Ω.
• The current in the 4.0 Ω resistor and in the 8.0 Ω resistor are 6 A and 3 A respectively.

Equivalent resistance of a parallel resistor circuit

The resulting or equivalent resistance is called the value of the resistance that is obtained by associating a set of them.

Resistors are said to be connected together in parallel when their two terminals are respectively connected to each terminal of the other resistor or resistors. In this way, in the parallel resistor circuit, the current is divided and circulates through several paths.

The equivalent resistance of a parallel resistor circuit is equal to the reciprocal of the sum of the inverses of the individual resistors. In the case of having 3 resistors then it is expressed as:

$R=\frac{1}{\frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} }$

Ohm's Law

Ohm's Law relates the magnitudes of voltage, resistance and intensity, expressing that the intensity of current that passes through a circuit is directly proportional to its voltage or voltage and inversely proportional to the resistance it presents.

Ohm's law is expressed mathematically as:

$I=\frac{V}{R}$

Where I is the intensity that is measured in amps (A), V the voltage that is measured in volts (V); and R is the resistance that is measured in ohms (Ω).

This case

A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel across a 24.0 V battery.

Being:

• R1= 4 Ω
• R2= 8 Ω
• R3= 12 Ω

the equivalent resistance of the circuit can be calculated as:

$R=\frac{1}{\frac{1}{4} +\frac{1}{8} +\frac{1}{12} }$

Solving:

R= 2.18  Ω

The equivalent resistance of the circuit is 2.18 Ω.

The current in the 4.0 Ω resistor and in the 8.0 Ω resistor is calculated using Ohm's law, considering that if two or more components are connected in parallel, they have the same potential difference (voltage) at their ends.

So, being V= 24 V, the corresponding currents are calculated as:

$I_{1} =\frac{V}{R1}$

$I_{1} =\frac{24 V}{4 ohm}$

$I_{1} =$ 6 A

and

$I_{2} =\frac{V}{R2}$

$I_{2} =\frac{24 V}{8 ohm}$

$I_{2} =$ 3 A

The current in the 4.0 Ω resistor and in the 8.0 Ω resistor are 6 A and 3 A respectively.

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