A parallel combination of two resistors, of 1Ω each, is connected in series with a 1.5Ω resistor. The total combination is connected across a 10V battery. The current flowing (in ampere) in the circuit is
Answers
Answer:
1Ω and 1Ω in parallel connection.
1/R eq = 1+1
= 2
R eq = 1/2Ω
1/2 Ω is in series with 1.5Ω
R eq = 1/2+1.5
= 0.5+1.5 = 2Ω
V = iR
given:
V = 10
R = 2Ω
by putting values
10 = 2i
i = 10/2
i = 5 Ampere (answer)
Answer:
Hence, the current that is flowing through the circuit = 5 A
Explanation:
In order to find the the current flowing in the circuit, we must, find the equivalent resistance of the circuit which will divide the potential difference.
Firstly we must calculate the resistance of resistors connected in parallel :
Given, resistors = 1Ω and 1Ω
We know that,
R(eq - P) = Equivalent resistance of parallel combination
1/R(eq - P) = 1/R1 + 1/R2
=> 1/R(eq) = 1/1 + 1/1
=> 1/R(eq) = 2/1
=> R(eq) = 1/2 Ω = 0.5 Ω
Hence, the resistance of the resistors connected in parallel is 0.5 Ω
Now, let us calculate the resistance of of the component which is made by connecting the above parallel component in series with the resistor of 1.5 Ω.
Given, resistors = 0.5Ω and 1.5Ω
We know that,
R(eq - S) = Equivalent resistance of series combination.
R(eq - S) = R1 + R2
=> R(eq) = 0.5 + 1.5
=> R(eq) = 2 Ω
Hence, the resistance of the combination is 2Ω.
So finally, we get, the total resistance of the circuit = 2 Ω.
Also,
Resistance, (R) = Potential Difference (V) / Current (I)
=> R = V / I
From this we can derive that,
=> I = V / R
Here,
V = 10 V
R = 2 Ω
By applying values in the formula, we get,
=> I = 10/2 = 5 A
Hence, the current that is flowing through the circuit = 5 A
NOTE : THE FIGURE OF CIRCUIT IS GIVEN IN ATTACHMENT BELOW.
