A P.D. of 2 volts exists across a potentiometer wire of length 4 m. When the P.D. across a `2 Omega` resistance of a second circuit is measured by this potentiometer, the balancing length is found to be 4 cm, the current in the second circuit is

Options

A. 1 mA
B. 10 mA
C. 100 mA
D. 50 mA

Answer

B

Explanation

Potential gradient `=(2)/(4)=(1)/(2) V//m`
In the second circuit, the balancing length =4 cm
`therefore P.D. = 4xx10^(-2)xx(1)/(2)=2xx10^(-2)V`
`therefore` Current in the second circuit
`= (2xx10^(-2)V)/(2 Omega)=10^(-2) A= 10 mA`