CUET Preparation Today
A combination of resistors is shown below. The equivalent resistance between the terminals P and Q is:
|
$\frac{11}{4}Ω$ $\frac{4}{11}Ω$ $7Ω$ $\frac{14}{5}Ω$ |
$\frac{11}{4}Ω$ |
The correct answer is Option (1) → $\frac{11}{4}Ω$
$\frac{1}{R_2}=1+\frac{1}{3}=\frac{4}{3}$ $⇒R_2=\frac{3}{4}$ $R_{net}=1+1+\frac{3}{4}=\frac{11}{4}Ω$ |
