The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its stand and cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l =30 cm from the positive end, the e.m.f. of the battery is (where i is the current in the potentiometer wire).

Question

The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its stand and cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l =30 cm from the positive end, the e.m.f. of the battery is (where i is the current in the potentiometer wire). (A) (30 E/100) (B) (30 E/100-0.5) (C) (30(E-0.5 i )/100) (D) (30 E/100 .5)

Tardigrade Admin · Accepted Answer

Let V be the potential across balance point and one end of wire. Hence acccording to the principle of potentiometer V propto l Also if a cell of emf E is employed in the circuit between the ends of potentiometer wire of length L then E propto L . Therefore, ( V / E )=(1/ L ) V =(1/ L ) E =( 3 0 / 1 0 0 ) E =( 3 0 E / 1 0 0 )

$\frac{30 E}{100}$

$\frac{30 E}{100 - 0.5}$

$\frac{30 ( E - 0.5 i )}{100}$

$\frac{30 E}{100.5}$

10030E​

100−0.530E​

10030(E−0.5i)​

100.530E​

Let V be the potential across balance point and one end of wire. Hence acccording to the principle of potentiometer V∝l Also if a cell of emf E is employed in the circuit between the ends of potentiometer wire of length L then E∝L. Therefore, EV​=L1​ V=L1​E=10030​E=10030E​

$\frac{30 E}{100}$

$\frac{30 E}{100 - 0.5}$

$\frac{30 ( E - 0.5 i )}{100}$

$\frac{30 E}{100.5}$