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Chapter 14: Problem 43

Kirchhoff's second law is based on the law of conservation of (A) Momentum (B) Charge (C) Energy (D) Sum of mass and energy

Short Answer

Expert verified
Kirchhoff's second law, also known as the Voltage Law or Loop Rule, is based on the law of conservation of energy. It states that the sum of voltage drops and sources in a closed loop of a circuit is equal to zero, which implies that the total energy in the loop is conserved. Hence, the correct answer is (C) Energy.

Step by step solution

01

Understand Kirchhoff's Second Law

Kirchhoff's second law, also known as the Voltage Law or Loop Rule, states that the sum of the voltage (electric potential) drops around a closed loop in a circuit is equal to the sum of the voltage sources present in that loop. In other words, the algebraic sum of all the potential differences (voltages) around any closed loop or mesh in a network is always equal to zero.
02

Relate Kirchhoff's Second Law to the Law of Conservation

Kirchhoff's second law is based on the law of conservation of energy. This is because the voltage around a closed loop in a circuit represents electric potential energy. When we sum up the voltage drops and sources in the loop and equate it to zero, it implies that the total energy in the loop is conserved.
03

Identify the Correct Option

From the analysis above, it is clear that Kirchhoff's second law is based on the law of conservation of energy. Therefore, the correct option is: (C) Energy

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Most popular questions from this chapter

Two conductors have the same resistance at \(0^{\circ} \mathrm{C}\) but their temperature coefficients of resistance are \(\alpha_{1}\) and \(\alpha_{2} .\) The respective temperature coefficients of their series and parallel combinations are nearly (A) \(\frac{\alpha_{1}+\alpha_{2}}{2}, \alpha_{1}+\alpha_{2}\) (B) \(\alpha_{1}+\alpha_{2}, \frac{\alpha_{1}+\alpha_{2}}{2}\) (C) \(\alpha_{1}+\alpha_{2}, \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}\) (D) \(\frac{\alpha_{1}+\alpha_{2}}{2}, \frac{\alpha_{1}+\alpha_{2}}{2}\)

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are \(3 \%\) each, then error in the value of resistance of the wire is (A) \(6 \%\) (B) Zero (C) \(1 \%\) (D) \(3 \%\)

Two cells with the same EMF \(E\) and different internal resistances \(r_{1}\) and \(r_{2}\) are connected in series to an external resistance \(R\). The value of \(R\) for the potential difference across the first cell to be zero is (A) \(\sqrt{r_{1} r_{2}}\) (B) \(r_{1}+r_{2}\) (C) \(r_{1}-r_{2}\) (D) \(\frac{r_{1}+r_{2}}{2}\)

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The balance length when key \(K_{1}\) is open and \(K_{2}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(25 \mathrm{~cm}\)

A potentiometer is a device used for measuring EMF and internal resistance of a cell. It consists of two circuits, one is main circuit in which there is a cell of given emf \(\varepsilon^{\prime}\) and given resistance \(R\) which is connected across a wire of length \(100 \mathrm{~cm}\) and having resistance \(r\) and another circuit having unknown EMF \(\varepsilon\) and galvanometer. For a given potentiometer, if \(\varepsilon^{\prime}=30 \mathrm{~V}, r=1 \Omega\), and resistance \(R\) varies with time \(t\) given by \(R=2 t\). The jockey can move on wire with constant velocity \(10 \mathrm{~cm} / \mathrm{s}\) and switch \(S\) is closed at \(t=0\)If jockey starts moving from \(A\) at \(t=0\) and balancing point found at \(t=1 \mathrm{~s}\) then the value of \(\varepsilon\) is (A) \(1 \mathrm{~V}\) (B) \(2 \mathrm{~V}\) (C) \(3 \mathrm{C}\) (D) \(4 \mathrm{~V}\)
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