Question

In The network shown in given figure,find the current I1 through the 10 V battery.​

Answer 1

Q]__?

Applying junction rule to C and D

ɪ = ɪ1 + ɪ2 _[1]

10 V battery is connected alone across CD so

Vᴅ - Vc = 10 V

∴I2 = Vᴄ - Vᴅ/ 2Ω

= -10 V/ 2Ω

= -5A

(Negative sign indicated current I2 is form D to C)

Applying loop rule to CDEFGC,

2l2 + 5l + 3l - 20 = 0

=> 2(-5) + 8l -20 = 0

=> -10 + 8l - 20 = 0

=> l = 30/8 = 15/4 A

From equation [1] we get,

I1 = l - l2 = 15/4 A- (-5A)

= [ 15/4 +5 ] A

= 35/4 A

∴ The Current I1 is 35/4 A

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Answer 2

Answer:

The current I1 is 35/4A...

Hope it will be helpful ☺️ thank you!