Question 74: Current Electricity

Question

If $$P=Q=R=10\Omega$$ and $$S=20\Omega$$, then what resistance should be joined with S to balance the Wheatstone's network?

Accepted Answer

The given Wheatstone bridge circuit is shown below.
Wheatstone is balanced only when
$$\frac{P}{Q}=\frac{R}{S}$$
Thus, the value of resistance S should be 10$$\Omega$$. But it is given to 20$$\Omega$$.
Hence, to decrease its value upto 10$$\Omega$$, suppose $$x ~\Omega$$ resistance is connected parallel to it.
$$	herefore \frac{1}{10}=\frac{1}{20}+\frac{1}{x}$$
$$\Rightarrow \frac{1}{x}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$$
$$\Rightarrow x=20~\Omega$$

Answer

Join a resistance of 10$$\Omega$$ in series with S

Answer

Join a resistance of 10$$\Omega$$ in parallel with S

Answer

Join a resistance of 20$$\Omega$$ in parallel with S

Answer

Join a resistance of 20$$\Omega$$ in series with S

Current Electricity

If $P=Q=R=10\Omega and S=20\Omega$, then what resistance should be joined with S to balance the Wheatstone's network?

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