Question

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Answer 1

Question :-

In the adjoining diagram , electromotive force of the cell is 4 V and the internal resistance is negligible. Find :

• Total Resistance

• Current of the Circuit

• Potential Difference between the point A and C.

To Find :-

• Total Resistance .

• Current of the Circuit.

• Potential Difference between the point A and C.

Given :-

• Electromive Force = 4 V

• Resistance between D and C = 3 ohm

• Resistance between F and E = 1 ohm

• Resistance between A and B = 3 ohm

We know :-

⠀⠀⠀⠀⠀Total Resistance in a parallel circuit :-

$\boxed{\underline{\bf{\dfrac{1}{R_{p}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + . . . + \dfrac{1}{R_{n}}}}}$

Where :-

• $R_{p}$ = Total Resistance in a parallel circuit.

• R = Resistance

⠀⠀⠀⠀⠀Total Resistance in a series circuit :-

$\boxed{\underline{\bf{R_{s} = R_{1} + R_{2} + R_{3} + . . . + R_{n}}}}$

Where :-

• $R_{p}$ = Total Resistance in a series circuit.

• R = Resistance

⠀⠀⠀⠀⠀Ohm's Law :-

$\boxed{\underline{\bf{V = IR}}}$

Where :-

• V = Potential Difference

• I = Current flowing through the conductor.

• R = Resistance offered.

Concept :-

In the second diagram , the figure was made by extending the wire !!

So we get the AC and BC in a Parallel Circuit , and CD in q Series one.

So by finding the resistance between AC and BC , by using the formula for Total Resistance in a Parallel Circuit.

And then by adding the Resistance between AB and BC and CD , by using the formula for Resistance in a series circuit , we can find the the Total Resistance in the circuit.

After Finding the total resistance we can use the ohms law to find the current in the circuit .

To find the potential between AC , we have to first find the current in the circuit and then also using the ohms law , we can find the required value.

Solution :-

⠀⠀⠀⠀⠀⠀Total Resistance :-

Resistance in the parallel circuit :-

Given :-

• $R_{1}$ = 3 ohm

• $R_{2}$ = 3 ohm

Using the formula and substituting the values in it , we get :-

$:\implies\bf{\dfrac{1}{R_{p}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}} \\ \\ \\ :\implies\bf{\dfrac{1}{R_{p}} = \dfrac{1}{3} + \dfrac{1}{3}} \\ \\ \\ :\implies\bf{\dfrac{1}{R_{p}} = \dfrac{1 + 1}{3}} \\ \\ \\ :\implies \bf{\dfrac{1}{R_{p}} = \dfrac{2}{3}} \\ \\ \\ :\implies\bf{R_{p} = \dfrac{3}{2}} \\ \\ \\ :\implies\bf{R_{p} = 1.5\:\Omega} \\ \\ \\ \therefore \purple{\bf{R_{p} = 1.5\:\Omega}}$

Hence, the resistance in the parallel circuit is 1.5 ohm.

Resistance in the series circuit :-

Given :-

• $R_{1}$ = 1.5 ohm
• $R_{2}$ = 1 ohm

Using the formula and substituting the values in it, we get :-

$:\implies \bf{R_{s} = R_{1} + R_{2}} \\ \\ \\ :\implies \bf{R_{s} = 1.5 + 1} \\ \\ \\ :\implies \bf{R_{s} = 2.5} \\ \\ \\ \therefore \purple{\bf{R_{s} = 2.5\:\Omega}}$

Hence, the resistance in the series circuit is 2.5 ohm.

But we have find the resistance by including all the resistors , so we Total Resistance in the circuit is 2.5 Ohm.

⠀⠀⠀⠀⠀⠀⠀⠀Current in the Circuit :-

Given :-

• Resistance = 2.5 ohm

• Potential Difference = 4 V

Using the formula and substituting the values in it, we get :-

$:\implies \bf{V = IR} \\ \\ \\ :\implies \bf{4 = I \times 2.5} \\ \\ \\ :\implies \bf{\dfrac{4}{2.5} = I} \\ \\ \\ :\implies \bf{1.6 A = I} \\ \\ \\ \therefore \purple{\bf{l = 1.6 A}}$

Hence, the current flowing through the conductor is 1.6 A.

⠀⠀⠀⠀Potential difference between A and B :-

Given :-

• R = 3 ohm

• Current = 1.6 A

(After stretching the wire , the circuit will become a series circuit and we know that in a series circuit the current is equally distributed.)

Now ,

Now using the formula and substituting the values in it , we get :-

$:\implies \bf{V = IR} \\ \\ \\ :\implies \bf{P = 1.6 \times 3} \\ \\ \\ :\implies \bf{P = 4.8} \\ \\ \ \therefore \purple{\bf{l = 4.8 V}}$

Hence, the potential difference between A and B is 4.8 V

Answer 2

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