All resistances in figure are 1 Ω each. The value of current 'I' is a/5 A. The value of a is _________. Numerical answer type. Work it out, then open the answer.

Answer: 8. Work from the right: last stage, two 1 Ω in parallel = 0.5 Ω. Next stage: (1 + 0.5) ∥ (1 + 0.5) = 0.75 Ω; then (1 + 0.75) ∥ (1 + 0.75) = 0.875 Ω. Total = 1 + 0.875 = 1.875 Ω I = (3)/(1.875) = 1.6 A = 8/5 A Answer: a = 8

All resistances in figure are 1 Ω each. The value of current 'I' is… | JEE Main 2022 (28 Jun, Shift 2)

Q115JEE Main 2022 · 28 JUN · SHIFT 2Difficult

Kirchhoff's Laws & Measuring Instruments

All resistances in figure are 1 $\Omega$ each. The value of current 'I' is $\frac{a}{5}$ A. The value of a is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

Work from the right: last stage, two $1\ \Omega$ in parallel $= 0.5\ \Omega$ .
Next stage: $(1 + 0.5) \parallel (1 + 0.5) = 0.75\ \Omega$ ; then $(1 + 0.75) \parallel (1 + 0.75) = 0.875\ \Omega$ .
Total $= 1 + 0.875 = 1.875\ \Omega$
$I = \frac{3}{1.875} = 1.6\text{ A} = \frac{8}{5}\text{ A}$
Answer: $a = 8$

Report a mistake

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2022 (28 Jun, Shift 2)

More questions on Kirchhoff's Laws & Measuring Instruments