Question

Two resistors of 2 ohm and x ohm are connected in series across a 12 V battery. If the voltage

drop across X ohm is 3 V. Find the Current across 2 Ohm resistor and the value of X.​

Answer 1

Answer:

• The current across 2 Ω resistor = 4.5 amperes
• The value of X = 2/3 Ω

Step-by-step explanation:

• When resistors are connected in series combination then, Rₛ = R₁ + R₂

Where,

• Rₛ = Equivalent resistance
• R₁ = Resistance of first resistor
• R₂ = Resistance of second resistor

We know that,

• I = V/R

Where,

• R = Resistance
• V = Potential difference
• I = Current

In series combination, current is same at each resistor wheres potential difference is different.

Given that:

• Two resistors of 2 Ω and X Ω are connected in series across a 12 V battery.
• R₁ = 2 Ω
• R₂ = X Ω
• V = 12 V
• I = V/(R₁ + R₂) ⇒ I = V/Rₛ

⟶ I = 12/(2 + X) _____(i)

• The voltage drop across X Ω is 3 V.
• R₂ = X Ω
• V₂ = 3 V

⟶ I = 3/X _____(ii)

To Find:

• The current across 2 Ω resistor and the value of X.

Finding the value of X:

Comparing eqⁿ(i) and eqⁿ(ii).

⇒ 12/(2 + X) = 3/X

⇒ 12 × X = 3(2 + X)

⇒ 12X = 6 + 3X

⇒ 9X = 6

⇒ X = 6/9

⇒ X = 2/3

∴ The value of X = 2/3 Ω

Finding the current across 2 Ω resistor:

In equation (ii).

⇒ I = 3/X

Substituting the value of X.

⇒ I = 3/(2/3)

⇒ I = (3 × 3)/2

⇒ I = 9/2

⇒ I = 4.5

∴ Current = 4.5

As we know in series combination current is same at each resistor.

Therefore, The current across 2 Ω resistor = 4.5 ampere

Answer 2

Answer:

Given :-

Two resistor of 2 Ω and x Ω

Potential difference = 12 V

Voltage drop across x Ω is 3 V

To Find :-

Current across 2 Ω

Value of x

Concept :-

Here we have to first find resistance by using the series combination

1/n = 1/R1 + 1/R2 + 1/R3

Let the two resistor be R1 and R2

$\sf \: R_1 = 2$

$\sf \: R_2 = x$

Therefore

V = IR

I = V/R

I = 12/(2 + x)

I = 12/2 + x

Again

I second one = 3/x

$\sf \dfrac{12}{2 + x} = \dfrac{3}{x}$

$\tt \: 12 \times x = 3(2 + x)$

$\tt \: 12x = 6 + 3x$

$\tt \: 12x - 3x = 6$

$\tt \: 9 x = 6$

$\tt \: x = \dfrac{6}{9}$

$\tt \: x = \dfrac{2}{3}$

Again

I = 3/(2/3)

I = 3 × 3/2

I = 9/2

I = 4.5 A