Answer:
Explanation:
Given that three resistor of resistance
R1 = 4 Ω
R2 = 6 Ω
R3 = 10 Ω
The resistor are connected in parallel, the equivalent resistance is
1 / Req = 1 / R1 + 1 / R2 + 1 / R3
Rearranging this we have
Req = R1•R2•R3 / (R2•R3 + R1•R3 + R1•R2)
Req = 4 × 6 × 10 / (6×10 + 4 × 10 + 4 × 6)
Req = 240 / (60 + 40 + 24)
Req = 240 / 124
Req = 1.94 Ω
The parallel connection is connected in series with a battery and a resistor.
Resistor resistance is
r = 2 Ω
Supply voltage of the battery is 12V
V = 12V
Let find the current flowing in the circuit.
V = I(Req + r)
I = V / (Req + r)
I = 12 / (1.94 + 2)
I = 12 / 3.94
I = 3.05A
So, this is the current flowing in the circuit and It is the same current that will be shared by the parallel resistance.
We can calculated the voltage across the parallel resistance
From ohms las
V = iR
V = I × Req
V = 3.05 × 1.94
V = 5.92 V.
Then, this the voltage across each parallel resistor.
Then, to know the current in the 10ohms resistance
V = iR
I = V / R3.
I = 5.92 / 10
I = 0.592 A.
The current in the 10 ohms resistor is 0.59A
