Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p=1…
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 :
0.9 A
1.35 A
0.3 A
1.0 A
Solution
The circuit can be considered as $\begin{aligned} & \therefore \mathrm{R}_{\text {eq }}=0.5+\frac{0.5 \times 2}{2+0.5}=\left(\frac{5}{10}+\frac{10}{25}\right) \Omega \\ & =\frac{45}{50}=\frac{9}{10}=0.9 \\ & \therefore \mathrm{i}=\frac{0.9}{0.9}=1 \mathrm{~A}\end{aligned}$
