since the circuit is connected parallel...
the equivalent resistance 1/R.eq=1/R1+1/R2+1/R3
1/R.eq=1/5+1/10+1/30
Taking Lcm we get:-
1/R.eq= 6+3+1
_____________
30
1/R.eq=10/30=1/3
therefore R.eq=3ohm
i) thus effective resistance is 3ohm
ii) current in ammeter= V/R (since V=IR)
I = 10/3= 3.33.....ampere
plz mark as BRAINLiST
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Verified answer
since the circuit is connected parallel...
the equivalent resistance 1/R.eq=1/R1+1/R2+1/R3
1/R.eq=1/5+1/10+1/30
Taking Lcm we get:-
1/R.eq= 6+3+1
_____________
30
1/R.eq=10/30=1/3
therefore R.eq=3ohm
i) thus effective resistance is 3ohm
ii) current in ammeter= V/R (since V=IR)
I = 10/3= 3.33.....ampere
plz mark as BRAINLiST