In The network shown in given figure,find the current I1 through the 10 V battery.
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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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The current I1 is 35/4A...
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