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Three resistors 5Ω, 10Ω and 30Ω are connected in parallel with the battery of Voltage 6V?
Calculate:
1. The value of current across each resistor.
2. The value of Potential difference across each resistor.
3. Total current in the circuit.
4. Effective resistance of the circuit
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Answers
Step-by-step explanation:
1.
Current through 5Ω resistor (I1) = V/R1
= 6/5
= 1.2 A
Current through 10Ω resistor (I2) = V/R2
= 6/10
= 0.6 A
Current through 30Ω resistor (I3)= V/R3
= 6/30
= 0.2 A
2.
Potential difference across each resistor is 6V, being parallel circuit.
3.
Total current in the circuit = I1 + I2 + I3
= 1.2 + 0.6 + 0.2
= 2 A
4.
Effective resistance of the circuit(Rp)
= 1/5 + 1/10 + 1/30
= 6+3+1/30
= 10/30 = 1/3
= 3Ω
Answer: 3Ω
Given:
Resistance = 5Ω, 10Ω and 30Ω
Voltage of battery = 6V
Find:
The value of current across each resistor.
The value of Potential difference across each resistor.
Total current in the circuit.
Effective resistance of the circuit.
Solution:
1)
we, know that
→ I = V/R
Current in resistor of 5Ω
⊙ I₁ = V/R
⊙ I₁ = 6/5 = 1.2 A
Current in resistor of 10Ω
⊙ I₂ = V/R
⊙ I₂ = 6/10 = 0.6 A
Current in resistor of 30Ω
⊙ I₃ = V/R
⊙ I₃ = 6/30
⊙ I₃ = 2/10 = 0.2 A
2)
♪ Potential difference across each resistor is 6V bcz it is a parallel circuit.
3)
we, use
→ 1/R = 1/R₁ + 1/R₂ + 1/R₃
→ 1/R = 1/5 + 1/10 + 1/30
→ 1/R = (6 + 3 + 1)/30
→ 1/R = 10/30
→ 1/R = 1/3
→ R = 3Ω
Total Current (I) = V/R
→ I = 6/3
→ I = 2A
4)
Total effective resistance will be 3Ω