Question 140: Current Electricity

Question

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p=1 \Omega$ as shown in the figure. An external resistance of $R_e=2 \Omega$ is connected via the sliding contact. The electric current in the circuit is :

Accepted Answer

We know, $$R = \int {{l \over A} \Rightarrow R \propto l} $$
As sliding contact of a potentiometer is in the middle of the potentiometer wire.
$	herefore$ $${R_{eq}} = 0.5 + {{0.5 	imes 2} \over {0.5 + 2}} = 0.5 + {1 \over {2.5}}$$
$$ = {1 \over 2} + {2 \over 5} = {{5 + 4} \over {10}} = {9 \over {10}} = 0.9\Omega $$
Hence, $$I = {v \over {{R_{eq}}}} = {{0.9v} \over {0.9\Omega }} \Rightarrow I = 1A$$

Answer

1.35 A

Answer

0.9 A

Answer

1.0 A

Answer

0.3 A

Current Electricity

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance R_p=1 \Omega as shown in the figure. An external resistance of R_e=2 \Omega is connected via the sliding contact. The electric current in the circuit is :

Choose the correct answer: