three resistors of 5ohm,10ohm and 30ohm are connected in parallel to a 6v battery. find
a) the value of current through each resistor
b) the total current in the circuit
c) and the total effective resistance of the circuit
Answers
Answer:
Explanation:
A: According to Ohm's law
V= IR
that implies
I = V/R
NOW; I ( 5 ohm ) = 6/5 = 1.2 A ...........(1)
I ( 10 ohm) = 6/10 = 0.6 A ........... (2)
I ( 30 ohm) = 6/30 = 0.2 A ............ (3)
B: the total current in the circuit is (1) + (2) +(3) ( total current is the sum of current across each resistor)
So , I ( total ) = 1.2 + 0.6 + 0.2 = 2 A
C: total resistance in parallel is
1/Re = 1/R1 + 1/R2 + 1/R3 ( R1 IS 5 ohms ,R2is 10 ohms , and R3 is 30 ohms )
That will give you
Re= 3 ohms
Pls mark as the brainliest
Answer:
A ] 5Ω = 1.2 Ampere , 10Ω = 0.6 Ampere , 30Ω = 0.2 Ampere
B ] I = 2 Ampere
C ] R = 3Ω
Explanation:
★ CURRENT ELECTRICITY ★
★ GIVEN ;
» 3 RESISTANCE = 5Ω , 10Ω and 30Ω connected in Parallel.
» EMF OF CELL = 6 VOLT
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A ] CURRENT THROUGH EACH RESISTOR .
» As ; all are connected in PARALLEL .. SO ; POTENTIAL DIFFERENCE ACROSS ALL RESISTORS IS SAME .
★ I¹ ( Current in 5Ω Resistor ) = V / R = 6 / 5 = 1.2 Ampere.
★ I² ( Current in 10Ω Resistor ) = V / R =
V / R = 6 / 10 = 0.6 Ampere
★ I³ ( Current in 30Ω Resistor ) = V / R =
V / R = 6 / 30 = 0.2 Ampere .
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B ] TOTAL CURRENT IN CIRCUIT.
» Equivalent Resistance of 3 Resistors in Parallel is ;
1 / R = 1 / R¹ + 1 / R² + 1 / R³
= 1 / 5 + 1 / 10 + 1 / 30
= [6 + 3 + 1 / 30 ]
1 / R = 10 / 30
★ R = 30 / 10 = 3Ω .
[ TOTAL CURRENT ( I ) = V / R = 6 / 3 = 2 AMPERE ] .
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C ] EFFECTIVE RESISTANCE
R = 3 Ω .
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