/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 The resistance of the series com... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 34

The resistance of the series combination of two resistances is \(S\). When they are joined in parallel, the total resistance is \(P\). If \(S=n P\), then the minimum possible value of \(n\) is (A) 4 (B) 3 (C) 2 (D) 1

Short Answer

Expert verified
The minimum possible value of \(n\) is (C) 2.

Step by step solution

01

For two resistances \(R_1\) and \(R_2\) in series, the total series resistance is given by: \(S = R_1 + R_2\) For two resistances \(R_1\) and \(R_2\) in parallel, the total parallel resistance is given by: \(\frac{1}{P} = \frac{1}{R_1} + \frac{1}{R_2}\) We are given the relation between \(S\) and \(P\) as: \(S = n P\) #Step 2: Solve for one resistance in terms of the other and the total resistance#

First, let's solve the series resistance equation for \(R_1\): \(R_1 = S - R_2\) Now, let's substitute this expression into the parallel resistance equation: \(\frac{1}{P} = \frac{1}{S - R_2} + \frac{1}{R_2}\) #Step 3: Setup inequality and solve for R_2#
02

Since we are looking for the minimum value for \(n\), we need to find the smallest possible value of \(\frac{S}{P}\). Therefore, we will set an inequality with respect to \(R_2\): \(\frac{1}{P} > \frac{1}{S - R_2} + \frac{1}{R_2}\) Now, cross-multiply and simplify: \(R_2(S - R_2) > (S - R_2)R_2\) Notice that both sides are equal. Since we are looking for the smallest possible value of \(n\), we can take the equality case where: \(R_2(S - R_2) = (S - R_2)R_2\) #Step 4: Solve for n#

Now we have: \(S - 2R_2 = 0\) From the relation between \(S\) and \(P\): \(S = n P\) Substituting the value of \(S\) from the previous relation: \(n P - 2R_2 = 0\) Since the resistances are positive, in this case, the smallest possible value for \(n\) is when \(R_1 = R_2\). We can use the above equation to find the value of \(n\): \(n = 2\) So, the minimum possible value of \(n\) is: (C) 2

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Most popular questions from this chapter

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 EMF of the cell \(E_{2}\) is (A) \(1 \mathrm{~V}\) (B) \(2 \mathrm{~V}\) (C) \(3 \mathrm{~V}\) (D) \(4 \mathrm{~V}\)

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) \(\frac{r_{1}+r_{2}}{2}\)

An ammeter is obtained by shunting a \(30 \Omega\) galvanometer with a \(30 \Omega\) resistance. What additional shunt should be connected across it to double the range? (A) \(15 \Omega\) (B) \(10 \Omega\) (C) \(5 \Omega\) (D) None of these

\(A, B\), and \(C\) are voltmeters of resistances \(R, 1.5 R\), and \(3 R\), respectively. When some potential difference is applied between \(X\) and \(Y\), the voltmeter readings are \(V_{A}, V_{B}\), and \(V_{C}\), respectively. (A) \(V_{A}=V_{B}=V_{C}\) (B) \(V_{A} \neq V_{B}=V_{C}\) (C) \(V_{A}=V_{B} \neq V_{C}\) (D) \(V_{B} \neq V_{A}=V_{C}\)

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 \(K_{2}\) is open and \(K_{1}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(13.5 \mathrm{~cm}\)
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