Question

(a) In the given circuit diagram (bottom figure), find the value of unknown resistance $S$, in the balancing condition of meter bridge.
(b) In the given circuit diagram (top figure) write the value of resultant resistance in between $X$ and $Y$ when key $K$ is
(i) Opened
(ii) Closed.

Answer 1

(a) For the given meter bridge, $\frac{R}{S}=\frac{40}{60}$

$⟹S=\frac{3R}{2}$

$=6\Omega$

(b) (i) When key is opened, the resistance of upper branch is $6\Omega +4\Omega =10\Omega$

When key is open, the resistance of lower branch is $3\Omega +12\Omega =15\Omega$

Hence the equivalent resistance=$\frac{10\times 15}{10+15}\Omega$

$=6\Omega$

(ii) When the key is closed, the resistance of two resistors in left connected in parallel=$\frac{6\times 3}{6+3}\Omega$

$=2\Omega$

The resistance of two resistors in right connected in parallel=$\frac{4\times 12}{4+12}\Omega$

$=3\Omega$

Hence equivalent resistance connected in series=$5\Omega$