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(i) Calculate the equivalent resistance
(ii) Find:
(a) Voltmeter reading
(b) Ammeter reading
(iii) Voltage drop by 'r'
(iv) P.d. across the battery
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Answer 1

$\Large{\underbrace{\underline{\sf Understanding\:the\: Question}}}$

Here in this question, concept of wheat stone bridge as well as internal resistance is used. We are given a circuit  and we are asked to find- total resistance, voltmeter and Ammeter reading, voltage drop by r and potential difference across battery. This could be a very complicated question if you don't know method to solve it. Firstly we will solve the wheatstone bridge to find total resistance and continue to solve the problem by step by step.

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Solution :-

The given wheatstone bridge is balanced, so the resistance between wire BC will not be considered.

Now we can see that all the resistors are of 4Ω, so equivalent resistance for wheat stone bridge is 4Ω.

Now resistors of 4Ω, 1Ω, 3Ω, and 6Ω are in series connection with wheatstone bridge. So total resistance of series connection can be expressed as::

⇒ Rₙ=R₁+R₂+R₃...

By substituting given values::

⇒ Rₙ=4Ω+4Ω+1Ω+6Ω+3Ω

⇒ Rₙ=18Ω

So the total resistance of the circuit is 18Ω.

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~Finding reading of Ammeter

⇒ ∈=IRₙ

⇒ 12v=I(18Ω)

⇒ 12v/18Ω=I

⇒ 0.6 Ampere=I

This is the current in the circuit.

We know that current remains same in series connection, so reading of Ammeter will be 0.6 Ampere.

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~Finding voltmeter reading

⇒ V=IRₓ

Since voltmeter is across 4Ω resistor, value of Rₓ will be 4Ω.

⇒ V=(0.6 A)(4Ω)

⇒ V=2.4v

This is the reading of voltmeter

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~Finding p.d.

We have formula for e.m.f.

⇒ ∈=I(R+r)

⇒ ∈=IR+Ir

Replace IR by Vₙ (p.d.)

⇒ 12v=Vₙ+(0.6 A)(6Ω)

⇒ 12v=Vₙ+3.6v

⇒ 12v-3.6v=Vₙ

⇒ 8.4v=Vₙ

This is the potential difference across terminal of battery.

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~Finding voltage dropped by r

Since the total emf of battery was 12v but it remained only 8.4 connected to circuit. It means that rest of the voltage is dropped by r.

This situation can be expressed as::

⇒ Potential drop by r=∈-Vₙ

⇒ Potential drop by r=12v-8.4v

⇒ Potential drop by r=3.6v

This is the potential dropped by r.

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★ Terms used by me in solution :-

• r=internal resistance

• Rₙ=R=Equivalent resistance

• V=voltmeter reading

• Vₙ=Voltage in circuit or across terminals of battery

• I=Reading of Ammeter

• ∈=emf

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Answer 2

Answer:

a\Large{\underbrace{\underline{\sf Understanding\:the\: Question}}}

UnderstandingtheQuestion

Here in this question, concept of wheat stone bridge as well as internal resistance is used. We are given a circuit and we are asked to find- total resistance, voltmeter and Ammeter reading, voltage drop by r and potential difference across battery. This could be a very complicated question if you don't know method to solve it. Firstly we will solve the wheatstone bridge to find total resistance and continue to solve the problem by step by step.

_______________________________

Solution :-

The given wheatstone bridge is balanced, so the resistance between wire BC will not be considered.

Now we can see that all the resistors are of 4Ω, so equivalent resistance for wheat stone bridge is 4Ω.

Now resistors of 4Ω, 1Ω, 3Ω, and 6Ω are in series connection with wheatstone bridge. So total resistance of series connection can be expressed as::

⇒ Rₙ=R₁+R₂+R₃...

By substituting given values::

⇒ Rₙ=4Ω+4Ω+1Ω+6Ω+3Ω

⇒ Rₙ=18Ω

So the total resistance of the circuit is 18Ω.

_______________________________

~Finding reading of Ammeter

⇒ ∈=IRₙ

⇒ 12v=I(18Ω)

⇒ 12v/18Ω=I

⇒ 0.6 Ampere=I

This is the current in the circuit.

We know that current remains same in series connection, so reading of Ammeter will be 0.6 Ampere.

_______________________________

~Finding voltmeter reading

⇒ V=IRₓ

Since voltmeter is across 4Ω resistor, value of Rₓ will be 4Ω.

⇒ V=(0.6 A)(4Ω)

⇒ V=2.4v

This is the reading of voltmeter

_______________________________

~Finding p.d.

We have formula for e.m.f.

⇒ ∈=I(R+r)

⇒ ∈=IR+Ir

Replace IR by Vₙ (p.d.)

⇒ 12v=Vₙ+(0.6 A)(6Ω)

⇒ 12v=Vₙ+3.6v

⇒ 12v-3.6v=Vₙ

⇒ 8.4v=Vₙ

This is the potential difference across terminal of battery.

_______________________________

~Finding voltage dropped by r

Since the total emf of battery was 12v but it remained only 8.4 connected to circuit. It means that rest of the voltage is dropped by r.

This situation can be expressed as::

⇒ Potential drop by r=∈-Vₙ

⇒ Potential drop by r=12v-8.4v

⇒ Potential drop by r=3.6v

This is the potential dropped by r.

_______________________________

★ Terms used by me in solution :-

r=internal resistance

Rₙ=R=Equivalent resistance

V=voltmeter reading

Vₙ=Voltage in circuit or across terminals of battery

I=Reading of Ammeter

∈=emf

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