Discussion Forum

In a balanced meter bridge, the resistance of the bridge wire is 0.1 $\Omega$ cm. Unknown resistance X is connected in left gap and 6 $\Omega$  in the right gap.  null point divides the wire in the ratio  2:3. find the current drawn  from the battery of 5 v having negligible resistance

Answer:

Explanation:

According to question

$\frac{l_{1}}{l_{2}}=\frac{2}{3}$            (given)

 For potentimeter,   $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}}=\frac{2}{3}$

$\frac{x}{6}=\frac{2}{3}$

$\Rightarrow$       x= 4 $\Omega$m^-1

 Total resistance of known and unknown resistance =6+4=10 $\Omega$

The resistance of wire is 0.1 x 100= 10 $\Omega$

 So, effective resistance R_eff is given by

 $\frac{1}{R_{eff}}=\frac{1}{10}+\frac{1}{10}$

 $\Rightarrow$   R_eff= 5 $\Omega$

 Hence  , current drawn from the battery is given by Ohm'sa law  , V=IR

 $\Rightarrow$      I= $\frac{V}{R}=\frac {5}{5}$ = 1 A