Question

find the current drawn from the battery by the network of four resistors shown in the figure​

Answer 1

Answer:

CURRENT DRAWN IS 0.23 AMPERE.

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Answer 2

Check the attachment for better explanation!!

Now from the attachment we see ;

➝ R₁ and R₂ and R₃ are in series combination. ☛ R(S)₁

➝ R₄ is in series combination. ☛ R(S)₂

Now we see that,

➝ R(S)₁ and R(S)₂ are in parallel combination.

Now we know that,

☛ 1/Rp= 1/R₁ + 1/R₂ + ......... + 1/R(n)

➝ 1/Rp = 1/(10 + 10 + 10) + 1/10

➝ 1/Rp = 1/30 + 1/10

➝ 1/Rp = (1 + 3)/30

➝ 1/Rp = 4/30

➝ 4Rp = 30

➝ Rp = 30/4

➝ Rp = 7.5Ω

Now using Ohm's law :-

☛ V = I * R

Here,

• V = 3v
• R = 7.5Ω

Now putting values we get,

➺ 3 = I * 7.5

➺ I = 3/7.5

➺ I = 0.4 A

∴ Current drawn from battery = 0.4 Amperes.