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Chapter 12: Problem 1819

Resistors \(P\) and \(Q\) connected in the gaps of the meter bridge. the balancing point is obtained \(1 / 3 \mathrm{~m}\) from the zero end. If a \(6 \Omega\) resistance is connected in series with \(\mathrm{p}\) the balance point shifts to \(2 / 3 \mathrm{~m}\) form same end. \(\mathrm{P}\) and \(\mathrm{Q}\) are. (A) 4,2 (B) 2,4 (C) both (a) and (b) (D) neither (a) nor (b)

Short Answer

Expert verified
The resistors P and Q have values of 2 ohms and 4 ohms, respectively. The correct answer is (B).

Step by step solution

01

Find the initial ratio of P and Q

When the resistors are connected alone, the balance point is obtained 1/3 m from the zero end. This means that the length of the bridge arm with resistor P is 1/3 m, and the length of the arm with resistor Q is 2/3 m. Let's write the ratio of resistances: \( \frac{P}{Q} = \frac{1/3}{2/3} = \frac{1}{2} \) Now we have the initial ratio of P and Q.
02

Find the new ratio of P and Q when a 6 ohm resistor is connected in series with P

When a 6 ohm resistor is connected in series with P, the balance point shifts to 2/3 m from the same end. This means that the length of the bridge arm with the combination of P and the 6 ohm resistor is 2/3 m, and the length of the arm with resistor Q is 1/3 m. Let's write the ratio of resistances with the new lengths: \( \frac{P + 6}{Q} = \frac{2/3}{1/3} = 2 \) Now we have the new ratio of P and Q when 6 ohms are added to P.
03

Solve the system of equations for P and Q

Now we have the following system of equations: \( \frac{P}{Q} = \frac{1}{2} \) \( \frac{P + 6}{Q} = 2 \) We can solve this system of equations by substitution or by solving both variables simultaneously. Let's try substitution: From the first equation, we find that: \( P = \frac{1}{2}Q \) Substituting this into the second equation: \( \frac{\frac{1}{2}Q + 6}{Q} = 2 \) Now we can solve for Q: \( \frac{Q + 12}{2Q} = 2 \) \( Q + 12 = 4Q \) \( 12 = 3Q \) \( Q = 4 \) Now, substitute this value of Q back into the first equation to find P: \( P = \frac{1}{2} (4) \) \( P = 2 \) So, the resistors P and Q have values of 2 ohms and 4 ohms, respectively. The correct answer is (B).

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