All resistances in figure are 1 - each- The value of current -I- is

R= 1 + [(1∥ 1+1)∥ (1∥1 + 1) + 1]∥[(1 ∥ 1 + 1)∥(1 ∥ 1 + 1) + 1] = 15/8Ω⇒ I = V/R = 1/15/8 = 8/15A ...

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Byju's Answer

Standard IX

Physics

Resistors in Series and Parallel

All resistanc...

Question

All resistances in figure are 1 $Ω$ each. The value of current " $I$ " is

\frac{1}{15} A

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\frac{2}{15} A

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\frac{4}{15} A

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\frac{8}{15} A

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Solution

The correct option is D $\frac{8}{15} A$
$R = 1 + [(1 ∥ 1 + 1) ∥ (1 ∥ 1 + 1) + 1] ∥ [(1 ∥ 1 + 1) ∥ (1 ∥ 1 + 1) + 1]$

$= \frac{15}{8} Ω \Rightarrow I = \frac{V}{R} = \frac{1}{15 / 8} = \frac{8}{15} A$

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All resistances in figure are 1 Ω each. The value of current "I" is

The correct option is D 815AR=1+[(1∥1+1)∥(1∥1+1)+1]∥[(1∥1+1)∥(1∥1+1)+1] =158Ω⇒I=VR=115/8=815A

All resistances in figure are 1 $Ω$ each. The value of current " $I$ " is

The correct option is D $\frac{8}{15} A$
$R = 1 + [(1 ∥ 1 + 1) ∥ (1 ∥ 1 + 1) + 1] ∥ [(1 ∥ 1 + 1) ∥ (1 ∥ 1 + 1) + 1]$

$= \frac{15}{8} Ω \Rightarrow I = \frac{V}{R} = \frac{1}{15 / 8} = \frac{8}{15} A$