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In the circuit shown in figure reading of voltmeter is\[{V_1}\]when only\[{S_1}\]is closed. The reading of voltmeter is\[{V_2}\]when only\[{S_2}\]is closed. The reading of voltmeter is\[{V_3}\]when both\[{S_1}\]and\[{S_2}\]are closed. Then,
A) \[{V_2} > {V_1} > {V_3}\]
B) \[{V_3} > {V_2} > {V_1}\]
C) \[{V_3} > {V_1} > {V_2}\]
D) \[{V_1} > {V_2} > {V_3}\]

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Last updated date: 13th Jun 2024
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Answer
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Hint:In case of the closed circuit, we use Kirchhoff's voltage law to find the potential flowing through the given closed circuit.
Complete step by step answer:
Kirchhoff’s law:
Kirchhoff’s voltage law states that “the algebraic sum of the product of current and resistance of the closed circuit is equal to the total EMF provided in the circuit. The symbol of the product of current and resistance depends upon the direction of the current flow in the circuit.
By applying this law we can solve the given problem

(i) The voltmeter reads the value\[{V_1}\]when only the switch\[{S_1}\]is closed. Applying Kirchhoff’s voltage law in the given circuit in which only the switch\[{S_1}\]is closed. Therefore,
\[ \Rightarrow 3IR + IR = E\]
\[ \Rightarrow 4IR = E\]
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\[\therefore I = \dfrac{E}{{4R}}\]
Now we can find the value of the potential through the resistance $3R$ is,

\[ \Rightarrow {V_1} = IR\]
\[ \Rightarrow {V_1} = \dfrac{E}{{4R}} \times 3R\]
\[\therefore {V_1} = \dfrac{3}{4}E\]--------(1)

(ii) The voltmeter reads the value\[{V_2}\]when only the switch\[{S_2}\]is closed. In this case, on applying Kirchhoff’s voltage law we get,
\[ \Rightarrow 6IR + IR = E\]
\[ \Rightarrow 7IR = E\]
\[\therefore I = \dfrac{E}{{7R}}\]

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Therefore the voltage through the resistance $6R$ is,
\[ \Rightarrow {V_2} = \dfrac{E}{{7R}} \times 6R\]
\[\therefore {V_2} = \dfrac{6}{7}E\] -------------- (2)
(iii) The voltmeter reads the value \[{V_3}\]when both switches are closed. Kirchhoff's law is applied in this case and the two resistances $3R$ and $6R$ are in parallel now. Hence the net resistance for the parallel loop is
\[ \Rightarrow \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3}{6}\]
\[ \Rightarrow 2R\]
The Kirchhoff’s law is applied in this case,
\[ \Rightarrow IR + 2IR = E\]
\[ \Rightarrow 3IR = E\]
\[\therefore I = \dfrac{E}{{3R}}\]
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The voltage flows when both switches are closed,
\[ \Rightarrow {V_3} = \dfrac{E}{{3R}} \times 2R\]
\[\therefore {V_3} = \dfrac{2}{3}E\] ------------- (3)
From (1), (2) & (3), we can say that the\[{V_2} > {V_1} > {V_3}\].

Hence option A is the correct answer.

Note:
Ohm’s law states that the potential difference across two points is directly proportional to the current flowing through the resistance $R$. Using Ohm’s law, we are able to find the value of the voltage across any resistance $R$. Kirchhoff's voltage law is used to find the value of the voltage in the closed circuit.