Question

Two resistance coils, P and Q are placed in the gaps of a metre bridge. A balance point is found when the movable contact touches the bridge wire at a distance of 35.5cm from the end joined Eo P. When the coil P is shunted with a resistance of 10ohms, the balance point is moved through a distance of 15.5cm . Find the value of resistance P and Q.

Answer 1

Answer on Question #73951, Physics / Other

Two resistance coils, P and Q are placed in the gaps of a metre bridge. A balance point is found when the movable contact touches the bridge wire at a distance of $35.5\mathrm{cm}$ from the end joined Eo P. When the coil P is shunted with a resistance of 10ohms, the balance point is moved through a distance of $15.5\mathrm{cm}$. Find the value of resistance P and Q.

Solution:

At balanced situation of bridge,

$$\frac {P}{Q} = \frac {R}{S}$$

or

$$\frac {P}{Q} = \frac {l}{100 - l}$$

where, $l$ is the balancing length.

In first case, $l = 35.5\mathrm{cm}$

$$\frac {P}{Q} = \frac {35.5}{100 - 35.5} = \frac {35.5}{64.5} = \frac {7.1}{12.9}$$

Thus,

$$Q = \frac {12.9}{7.1} P$$

A shunt is a resistance connected in parallel arrangement with another resistance.

Shunt resistances follow the rule

$$\frac {1}{P} + \frac {1}{10} = \frac {1}{P _ {2}}$$

So,

$$P _ {2} = \frac {10P}{10 + P}$$

In this case,

$$\frac {P _ {2}}{Q} = \frac {35.5 - 15.5}{100 - 35.5 + 15.5} = \frac {20}{80} = 0.25$$ $$\frac {10P}{10 + P} = 0.25Q$$ $$\frac {10P}{10 + P} = 0.25 \cdot \frac {12.9}{7.1} P$$ $$10 + P = \frac {10 \cdot 7.1}{0.25 \cdot 12.9} = 22.02$$ $$P = 22.02 - 10 = 12.02 \mathrm{Ohm} \approx 12.0 \mathrm{Ohm}$$ $$Q = \frac {12.9}{7.1} \cdot 12.02 = 21.84 \mathrm{Ohm} \approx 21.8 \mathrm{Ohm}$$

Answer: $P = 12.0$ Ohm; $Q = 21.8$ Ohm.

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