In the circuit , the value of I in ampere is
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Heya!!
Since resistance across AD =
4&4 are in parallel nd in series with 4 ohm resistance.
= R = 4 + 2
= 6 ohms.
Then this is parallel with 4 ohms(BC) ..
= 1/R = 1/6 + 1/4
= 1/R = 5/12
= R = 2.4 ohms
This is in series with 1.6 ohm.. Hence total resistance = 2.4 + 1.6
= 4 ohms.
Since potential diff is same in parallel. So PD = 4 V
Since I = V/R
= 4/4 A
= 1 A
Suppose current in BC = i & current in AD = I.
Since i + I = 1 A,
4i = 6I
=) i = 3l/2
=) 3I/2 + I = 1
=) 5I/2 = 1
=) I = 2/5 A
= 0.4 A
Hope it helps uh!!
Since resistance across AD =
4&4 are in parallel nd in series with 4 ohm resistance.
= R = 4 + 2
= 6 ohms.
Then this is parallel with 4 ohms(BC) ..
= 1/R = 1/6 + 1/4
= 1/R = 5/12
= R = 2.4 ohms
This is in series with 1.6 ohm.. Hence total resistance = 2.4 + 1.6
= 4 ohms.
Since potential diff is same in parallel. So PD = 4 V
Since I = V/R
= 4/4 A
= 1 A
Suppose current in BC = i & current in AD = I.
Since i + I = 1 A,
4i = 6I
=) i = 3l/2
=) 3I/2 + I = 1
=) 5I/2 = 1
=) I = 2/5 A
= 0.4 A
Hope it helps uh!!
Anonymous:
bro, is it correct?
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