find current drawn from a battery by the network of 4 resistors shown in figure
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25
aloha!
>>----------✪-----------<<
there are four resistors of 10 ohm each
two of them in series
Req= R1 + R2
= 10 + 10
= 20 ohm-----------(1)
now again
Req= R1 + R2
= 10 + 10
= 20 ohm------------(2)
now, (1) and (2)
are in parallel
1/Req = 1/20 + 1/20
1/Req= 2/20
1/Req= 1/10
Req= 10 ohm.
now, the current v=IR
3= I × 10
current = 3/10
current = 0.3 A
>>----------✪-----------<<
hope it helps :^)
>>----------✪-----------<<
there are four resistors of 10 ohm each
two of them in series
Req= R1 + R2
= 10 + 10
= 20 ohm-----------(1)
now again
Req= R1 + R2
= 10 + 10
= 20 ohm------------(2)
now, (1) and (2)
are in parallel
1/Req = 1/20 + 1/20
1/Req= 2/20
1/Req= 1/10
Req= 10 ohm.
now, the current v=IR
3= I × 10
current = 3/10
current = 0.3 A
>>----------✪-----------<<
hope it helps :^)
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mehak113:
pzz
tell it with figure discription
see sissy first look at your diagram
in the right, the resistors are in series
in the left, the resistors are also in series.
so we combine the resistors as in a series connection.
ok
now both combined 20 ohm resistors are in parallel so, we will
use the equivalent resistance for resistors in parallel
thnks
Answered by
12
I hope this helps you
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