Question

a battery of 6v is connected in series with 1ohm 2ohm 3ohm and 6ohm resistors find the current through 3ohm resistors ​

Answer 1

Answer:

How precise do you want the answer? ;)

The first-order model is an ideal, 6 V battery connected to an ideal, 3 Ω resistor. Furthermore we would assume the wires are ideal (0 Ω). The current would simply be I = V/R = 6/3 = 2.000 A.

A second-order model would include the nominal source resistance of the battery and the resistance of the wires. Let’s assume the nominal source resistance of the battery is 0.1 Ω and the total resistance of the wires is 0.01 Ω. The current would then be I = V/R = 6/(3 + 0.1 + 0.01) = 1.929 A.

A third order model would take into account the temperature coefficient of the resistor and wires. (The resistance of wires, and of most resistors, will increase as power dissipation increases.) In addition, a third order model would take into account that the battery voltage will decrease as current increases, and the model will use the battery’s actual source resistance at this current level. So let’s assume that, after the resistor is connected to the battery, the resistance of the resistor goes up to 3.05 Ω, the battery voltage decreases to 5.91 V, and the resistance of the wiring increases to 0.015 Ω. We will assume the actual source resistance of the battery is 0.08 Ω at this current level. The current is now I = V/R = 5.91/(3.05 + 0.08 + 0.015 ) = 1.879 A.

A fourth-order model would include the uncertainty of each value in the equation. Propagation of Uncertainty Analysis would be used to determine the uncertainty of the final current value.

A fifth-order model would include string theory, dark energy, gravity waves, and the Heisenberg Uncertainty Principle. :)

Answer 2

Answer:

$\bigstar{\bold{Current\:=\:0.5\:A}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Voltage (V) = 6 V
• Resistance (R₁) = 1 Ω
• Resistance (R₂) = 2 Ω
• Resistance (R₃) = 3 Ω
• Resistance (R₄) = 6 Ω

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Current through the 3 Ω resistor (I)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Equivalent resistance of resistors in series is given by

  R = R₁ + R₂ + R₃ + R₄

→ Substituting the datas, we get to total resistance

 R = 1 + 2 + 3 + 6

R = 12 Ω

→ In a series combination current across each resistor is same.

→ By Ohm's law current flowing through a circuit is given by

  I = V/R

→ Substitute the datas, we get

  I = 6/12

  I = 0.5 A

$\boxed{\bold{Current\:=\:0.5\:A}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ In a series connection, current across each resistor is different, but voltage across ends of a conductor is different for the resistors.

→ In a parallel connection, voltage is same for the each resistor, but current flowing through the resistors is different.

→ Ohm's law state that the current flowing through a conductor is directly proportional to potential difference applied across its ends provided the temperature and other physical  conditions remain unchanged.

   V = RI

→ The unit of current is Ampere (A).