Physics, asked by israrkhan6023, 4 months ago

A cell of internal resistant 2.0 ohm and electromotive force (e.m.f.) 1.5 V is connected to a resistor of resistance 3.0 ohm what is the potential difference across 3ohm resistor. .(a) 5 V (b) 1.2 V (c) 0.9 V (d) 0.6 V

Answers

Answered by Anonymous
34

Answer :

  • The potential difference across 3 Ω resistor is 0.9 V.

Explanation :

Given :

  • Internal resistance of the cell, r = 2 Ω
  • Electromotive force (e.m.f) of the cell,ε = 1.5 V
  • Resistance in the resistor, R = 3 Ω

To find :

  • Potential difference across the 3 Ω resistor, V = ?

Knowledge required :

⠀⠀⠀⠀⠀⠀⠀⠀⠀● Formula for Electromotive force (e.m.f) :

⠀⠀⠀⠀⠀⠀⠀⠀⠀ε = l(R + r)

Where,

  • ε = Electromotive force
  • l = Current
  • R = Load resistance
  • r = Internal resistance

⠀⠀⠀⠀⠀⠀⠀⠀⠀ Ohm's law :

⠀⠀⠀⠀⠀⠀⠀⠀⠀V = IR

Where,

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

[Note : In a series circuit, the current passing through each resistor is equivalent to each other]

Solution :

First let us find the current in the circuit :

Using the formula for e.m.f and by substituting the values in it, we get :

⠀⠀⠀=> ε = l(R + r)

⠀⠀⠀=> 1.5 = I(3 + 2)

⠀⠀⠀=> 1.5 = I × 5

⠀⠀⠀=> 1.5/5 = l

⠀⠀⠀=> 0.3 = I

⠀⠀⠀⠀⠀∴ l = 0.3 A

Hence the current in the circuit is 0.3 A.

Now,

Let's find out the potential difference in the 3 Ω resistor:

By using the ohm's law and Substituting the values in it, we get :

⠀⠀⠀=> V = IR

⠀⠀⠀=> V = 0.3 × 3

⠀⠀⠀=> V = 0.9

⠀⠀⠀⠀⠀∴ V = 0.9 V

Therefore,

  • Potential difference across the 3 Ω resistor, V = 0.9 V.
Answered by Anonymous
38

Answer:

 \huge \bf \: required \: answer

Firstly we will see the formula.

   \huge \sf \:  e \:  = i(R+ r)

Here,

E = Electromagnetic force

I = Current

R = Load resistance

r = Internal Resistance

Now,

Ohm's law

 \huge \bf \: v \:  = ir

V = Potential difference

I = Current

R = Resistance

Now,

Let's solve

 \sf \implies \: e \:  = i(R + r)

 \sf \implies \: 1.5 = i(2 + 3)

 \sf \implies \: 1.5 = 5 \times i

  \sf \implies \:  \frac{1.5}{5}  = i

 \sf \implies \: i \:  = 0.3

 \huge \fbox {current \:  = 0.3 a}

Let's find potential difference

We will apply Ohm's law

 \sf \implies \: v \:  = i \times r

 \sf \implies \: v \:  = 0.3 \times 3

 \sf \implies \: v \:  = 0.9

Therefore, potential difference across 3ohm resistor is 0.9 Volt

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