When 2 resistors of 6Ω and 12Ω are connected to a battery of 6 v in series combination, what will be the current passing through the circuit?
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When 2 resistors of 6Ω and 12Ω are connected to a battery of 6 v in series combination, what will be the current passing through the circuit?
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ANSWER
Total current I flowing through the circuit I=
R
V
Here R=equivalentresistance=3.03Ω-----refer the figure given.
I
total
=
3.03
5
=1.65A
Since the resistors are in series connection (fig. B) current I will flow equally throughout the circuit.
Now calculating for voltage in resistor of 2Ω
V
B
=IR=3.30V, then current I flowing through 6Ω resistor I
6
Ω
=
6
3.30
=0.55
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Answer:
6+12=18ohm
I=R/V
l=18/6V
I=3ohm meter