if the total resistance of the given circuit is 20ohm,then calculate the resistance of a resistor X and current flowing through the circuit.
Answers
Given that the total resistance of a circuit is 20 ohm and is connected to a 6V battery.
Observing figure, we can say that 10 ohm and X are connected in series. Also 25 ohm and 75 ohm are connected in series and their resultant is connected in parallel.
We have to find the current in the circuit.
Take R1 = 10 ohm, R2 = X ohm, R3 = 25 ohm and R4 = 75 ohm.
Rs = R1 + R2
Rs = (10 + X) ohm
Also,
Rs' = R3 + R4
Rs' = (25 + 75) ohm
Rs' = 100 ohm
Resultant of Rs and Rs' are connected in parallel. So,
1/Rp = 1/Rs + 1/Rs'
(Rp = 20 ohm given in question)
1/20 = 1/(10 + X) + 1/100
1/(10 + X) = 1/20 - 1/100
1/(10 + X) = 1/20(1/1 - 1/5)
1/(10 + X) = 1/20(4/5)
1/(10 + X) = 1/25
Cross-multiply them,
10 + X = 25
X = 15
Therefore, the value of X is 15 ohm.
Also,
V = IR
Substitute the known values,
6 = I × 15
6/15 = I
0.4 = I
Therefore, the current in the circuit is 0.4 A.
Given:
- Total resistance in the given circuit = 20Ω
- Connected to 6V battery
To find:
- Resistance of resistor x
- Current flowing through the circuit
Solution:
Given, total resistance that means Rp = 20Ω
As , from observing by the figure 2 resistors ,
one resistor of 10Ω & other resistor of X Ω . The both resistors are connected in series
Let
- Resistor 1 be R1 = 10Ω
- Resistor 2 be R2 = X
- Resistor 3 be R3 = 25Ω
- Resistor 4 be R4 = 75Ω
As , we know that if two resistors are connected in series :
Rs = R1 + R2
→ Rs = 10 + X
_____________
Rs' = R3 + R4
→ Rs' = 25 + 75
→ Rs' = 100Ω
Since, Rs and Rs' are connected in parallel so,
1/Rp = 1/Rs + 1/Rs'
→ 1/Rp = 1/(10 + X) + 1/100
→ 1/20 = 1/(10 + X) + 1/100
→ 1/20 - 1/100 = 1/(10 + X)
→ (5 - 1)/100 = 1/(10 + X)
→ 4/100 = 1/10 + x
→ 1/25 = 1/(10 + X)
By cross multiplying
→ 25 = [ 10 + x ]
→ 25 - 10 = x
→ x = 15 Ω
Now, finding current flowing
V = I × R
Simplifying
I = V/R
♦ I = 6/15
♦ I = 0.4 A
Hence, current flowing through the circuit = 0.4 A