Find the current flowing in the 3ohms resistors for the network also the p. d across ten ohms and 2 ohms

2.72 A

7.41 V

3.95 V

Explanation:

Let's say I₁ is the current in the 3Ω resistor, I₂ is the current in the 10Ω resistor, and I₃ is the current in the 2Ω resistor.

Apply Kirchoff's voltage law to the top loop:

20 − 6 I₂ − 10 I₂ − 3 I₁ = 0

20 − 16 I₂ − 3 I₁ = 0

Apply Kirchoff's voltage law to the bottom loop:

20 − 4 I₃ − 2 I₃ − 3 I₁ = 0

20 − 6 I₃ − 3 I₁ = 0

Apply Kirchoff's current law at either junction:

I₁ − I₂ − I₃ = 0

I₁ = I₂ + I₃

Three equations, three variables.  To solve, let's multiply the first equation by 3 and the second equation by 8:

60 − 48 I₂ − 9 I₁ = 0

160 − 48 I₃ − 24 I₁ = 0

Add:

220 − 48 I₂ − 48 I₃ − 33 I₁ = 0

220 − 48 (I₂ + I₃) − 33 I₁ = 0

Substitute and solve:

220 − 48 I₁ − 33 I₁ = 0

220 − 81 I₁ = 0

I₁ = 2.72

Therefore, from the first equation:

I₂ = 0.741

And from either the second or third equation:

I₃ = 1.98

So the voltage across the 10Ω resistor is:

V = IR

V = (0.741 A) (10 Ω)

V = 7.41 V

And the voltage across the 2Ω resistor is:

V = IR

V = (1.98 A) (2 Ω)

V = 3.95 V

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answer:

yes

explanation: