Question

Resistance of a uniform wire is 6 Ω. It is cut into 3 equal pieces and form a

series combination then connected to a battery of 10 V. Find net current in

the circuit.​

Answer 1

Given

• Resistance of a wire = 6Ω
• The wire is cut into 3 equal pieces
• They are connected in series
• Potential Difference = 10 V

To Find

• Net Current

Solution

● Rₙₑₜ = R₁ + R₂ .... + Rₙ [Series Connection]

● V = IR [Ohm's Law]

✭ Net Resistance :

→ Rₙₑₜ = R₁ + R₂ .... + Rₙ

→ Rₙₑₜ = 2 + 2 + 2

→ Rₙₑₜ = 6Ω

✭ Current accross the circuit :

→ V = IR

→ 10 = I × 6

→ 10/6 = I

→ Current = 1.7 A

Answer 2

Answer:

Given :-

• Resistance of wire = 6 ohm
• Total pieces = 3
• Potential difference = 10 V

To Find :-

Net current

Answer :-

$\sf {R}_{net} = R_{1} + {R}_{2} .. {R}_{n}$

As we know that

6/3 = 2

$\sf \: R_{net} = 2 + 2 + 2$

$\sf \:{ R}_{net} = 6 ohm$

Now,

Let's find current accross

Here,

We will use Ohm law

V = IR

10 = I × 6

10 = 6I

I = 10/6

I = 1.7 A

Hence :-

Current = 1.7 A