Question

in the following circuit, a meter bridge is shown in its balanced state. the bridge wire has a resistance of 1 ohm/cm. calculate the value of the unknown resistance X connected in the left gap when 3-ohm resistor is connected in the right gap and the current drawn from the battery of negligible internal resistance. [2ohm , 1.26A]

Answer 1

using principle of Wheatstone bridge,

we have,

$\frac{X}{40}=\frac{3}{60}$

$\Rightarrow X=2\Omega$

Combination of resistances

$=2+3=5\Omega$

resistance of the bridge wire,

$=100\times 1=100\Omega$

Total resistance,

$=\frac{100\times 5}{105}=\frac{500}{105}\Omega$

$I=\frac{V}{R}$

$=\frac{6}{\frac{500}{105}}$

$=1.26A$