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Current Electricity

Question

Wired Faculty App

In the circuit shown the value of I in ampere is

1

0.60

0.4

1.5

Solution

C.

0.4

We can simplify the network as shown

So net resistance,
R = 2.4 + 1.6 = 4.0 Ω

Therefore, current from the battery,

$i = \frac{V}{R}$

= $\frac{4}{4}$

i = 1A

Now from the circuit (b),

4 I' = 6I

⇒ I' = $\frac{3}{2}$ I

But, i = I + I'

= I + $\frac{3}{2}$ I

1 = $\frac{5}{2}$ I

⇒ I = $\frac{2}{5}$

I = 0.4 A

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