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1) as the resistors are connected in series ..the voltage drop across them will add up to 12V
but since 6V is the drop in unknown
so, drop in 3ohm =12-6=6V
2)as it's a series connection , so current will be same throughout the circuit
so, current through 3ohm resistance=
6/3=2A; {as V=IR}
so, current through unknown resistance is also 2A.
3) as V=IR and for the unknown resistance,
V=6V, I=2A
so, R=V/I=6/2=3ohm
Thanks.
but since 6V is the drop in unknown
so, drop in 3ohm =12-6=6V
2)as it's a series connection , so current will be same throughout the circuit
so, current through 3ohm resistance=
6/3=2A; {as V=IR}
so, current through unknown resistance is also 2A.
3) as V=IR and for the unknown resistance,
V=6V, I=2A
so, R=V/I=6/2=3ohm
Thanks.
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