Question

Q10. For a given circuit diagram
Resistors in Parallel
R,
w
R1 = 1 ohm, R2 = 2 ohm , R3 = 5 ohm
Battery = 6V
Calculate
[A] the current through the each resistor
[b] total current in the circuit
[c] total effective resistance of the circuit​

Answer 1

Answer:

Current through first resistor = 6 A

Current through second resistor = 3 A

Current through third resistor = 1.2 A

Tota current in the circuit = 10.2 A

Total effective resistance = 0.59 Ω

Explanation:

Given:

• R₁ = 1 Ω
• R₂ = 2 Ω
• R₃ = 5 Ω
• Potential difference = 6 V

To Find:

• Current through each resisitor
• Total current in the circuit
• Total effective resistance of the circuit

Solution:

Current through first resistor:

By Ohm's law we know that,

I = V/R

Substituting the data we get,

I  = 6/1

I₁ = 6 A

Hence current through the first resistor is 6 A.

Current through second resistor:

I₂ = V/R₂

Substitute the data,

I₂ = 6/2

I₂ = 3 A

Hence current through the second resistor is 3 A.

Current through the third resistor:

I₃ = V/R₃

Substitute the data,

I₃ = 6/5

I₃ = 1.2 A

Hence current through the third resistor is 1.2 A.

Total current in the circuit:

The total current in the circuit is the sum of the current flowing through the individual resistors.

I = I₁ + I₂ + I₃

Substitute the given data,

I = 6 + 3 + 1.2

I = 10.2 A

Hence the total current flowing through the circuit is 10.2 A.

Total effective resistance:

It is given that the resistors are conected in parallel combination.

Hence effective resistance is given by,

$\sf{\dfrac{1}{R}=\dfrac{1}{R_1} +\dfrac{1}{R_2} +\dfrac{1}{R_3} }$

Substitute the data,

$\sf{\dfrac{1}{R}=\dfrac{1}{1} +\dfrac{1}{2}+\dfrac{1}{5}}$

$\sf{\dfrac{1}{R}=\dfrac{10+5+2}{10}}$

$\sf{\dfrac{1}{R}=\dfrac{17}{10} }$

R = 10/17

R = 0.59 Ω

Hence the effective resistance of the circuit is 0.59 Ω.